Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

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Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

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Samacheer Kalvi 11th Chemistry Chapter 1 Basic Concepts of Chemistry and Chemical Calculations Textual Evaluation Solved

I. Choose the Best Answer

Samacheer Kalvi 11th Chemistry Chapter 1 Solutions Question 1.
40 ml of methane is completely burnt using 80 ml of oxygen at room temperature. The volume of gas left after cooling to room temperature is ………..
(a) 40 ml CO2 gas
(b) 40 ml CO2 gas and 80 ml H2O gas
(c) 60 ml CO2 gas and 60 ml H2O gas
(d) 120 ml CO2 gas
Answer:
(a) 40 ml CO2 gas
Solution:
CH4(g)  + 2O2(g)  CO2(g) + 2 H2O(l)

Content CH4 O2 CO2
Stoichiometric coefficient 1 2 1
Volume of reactants allowed to react 40 mL 80 mL          –
Volume of reactant reacted and product formed 40 mL 80 mL 40 mL
Volume of gas after cooling to the room temperature

Since the product was cooled to room temperature, water exists mostly as liquid. Hence, option (a) is correct

Basic Concepts Of Chemistry And Chemical Calculations Question 2.
An element X has the following isotopic composition 200X = 90 %, 199X = 8 % and 202X = 2 %. The weighted average atomic mass of the element X is closest to …………
(a) 201 u
(b) 202 u
(c) 199 u
(d) 200 u
Answer:
(d) 200 u
= \(\frac {(200 × 90) + (199 × 8) + (202 × 2) }{100}\) = 199.96 = 200 u

Basic Concepts Of Chemistry And Chemical Calculations Book Back Answers Question 3.
Assertion:
Two mole of glucose contains 12.044 × 1023 molecules of glucose.

Reason:
Total number of entities present in one mole of any substance is equal to 6.02 × 1022
(a) both assertion and reason are true and the reason is the correct explanation of assertion
(b) both assertion and reason are true but reason is not the correct explanation of assertion
(c) assertion is true but reason is false
(d) both assertion and reason are false
Answer:
(c) assertion is true but reason is false

Correct reason:
Total number of entities present in one mole of any substance is equal to 6.022 x 1023

Samacheer Kalvi Guru 11th Chemistry Question 4.
Carbon forms two oxides, namely carbon monoxide and carbon dioxide. The equivalent mass of which element remains constant?
(a) Carbon
(b) Oxygen
(c) Both carbon and oxygen
(d) Neither carbon nor oxygen
Answer:
(b) Oxygen

Reaction 1:
2 C + O2 → 2 CO2
2 × 12 g carbon combines with 32 g of oxygen.
Hence, Equivalent mass of carbon = \(\frac {2 × 12}{32}\) × 8 = 6

Reaction 2:
C + O2 → 2 CO2
12 g carbon combines with 32 g of oxygen.
Hence, Equivalent mass of carbon = \(\frac {12}{32}\) × 8 = 3

11th Chemistry Samacheer Kalvi Question 5.
The equivalent mass of a trivalent metal element is 9 g eq-1 the molar mass of its anhydrous oxide is ………..
(a) 102 g
(b) 27 g
(c) 270 g
(d) 78 g
Answer:
(a) 102 g
Let the trivalent metal be M3+
Equivalent mass = mass of the metal / valance factor
9g eq-1 = mass of the metal / 3 eq
Mass of the metal = 27 g
Oxide formed M2O3
Mass of the oxide = (2 × 27) + (3 × 16) = 102 g

11th Chemistry Lesson 1 Question 6.
The number of water molecules in a drop of water weighing 0.018 g is ………….
(a) 6.022 × 1026
(b) 6.022 × 1023
(c) 6.022 × 1020
(d) 99 × 1022
Answer:
(c) 6.022 × 1020
Weight of the water drop = 0.0 18 g
No. of moles of water in the drop = Mass of water / molar mass = 0.018/18 = 10-3 mole
No of water molecules present ¡n I mole of water = 6.022 × 1023
“No. water molecules in one drop of water (10 mole) = 6.022 × 1023 × 10-3
= 6.022 × 1020

Chemistry Class 11 Samacheer Kalvi Question 7.
1 g of an impure sample of magnesium carbonate (containing no thermally decomposable impurities) on complete thermal decomposition gave 0.44 g of carbon dioxide gas. The percentage of impurity in the sample is ………..
(a) 0 %
(b) 4.4 %
(c) 16 %
(d) 8.4 %
Answer:
(c) 16%
Mg CO3 → MgO + CO2
Mg CO3 : (1 × 24) + (1 × 12) + (3 × 16) = 84 g
CO2 : (1 × 12) + (2 × 16) 44g
100% pure 84 g MgCO3 on heating gives 44 g CO2
Given that I g of MgCO3 on heating gives 0.44 g CO2
Therefore, 84 g MgCO3 sample on heating gives 36.96 g CO2 = 100%
Percentage of purity of the sample = \(\frac{100 \%}{44 \mathrm{g} \mathrm{CO}_{2}}\) × 36.96 g CO2 = 84%
Percentage of impurity = 16%

Samacheer Kalvi 11th Chemistry Question 8.
When 6.3 g of sodium bicarbonate is added to 30 g of acetic acid solution, the residual solution is found to weigh 33 g. The number of moles of carbon dioxide released in the reaction is –
(a) 3
(b) 0.75
(c) 0.075
(a) 0.3
Answer:
(c) 0.075
Samacheer Kalvi 11th Chemistry Chapter 1 Solutions Basic Concepts Of Chemistry And Chemical Calculations
The amount of CO2 released, x = 3.3 g
No. of moles of CO2 released = 3.3 / 44 = 0.075 mol

Samacheer Kalvi 11th Chemistry Solutions Question 9.
When 22.4 liters of H2 (g) is mixed with 11.2 liters of Cl2 (g), each at 273 K at 1 atm the moles of HCl (g), formed is equal to ………..
(a) 2 moles of HCl (g)
(b) 0.5 moles of HCl (g)
(c) 1.5 moles of HCl (g)
(d) 1 moles of HCl (g)
Answer:
(d) 1 moles of HCl (g)
Solution:
H2(g) + Cl2(g) → 2 HCl (g)

Content H2(g) cl2(g) HCl (g)
Stoichiometric coefficient 1 1 2
No. of moles of reactants allowed to react at 273 K and 1 atm pressure 22.4 L (1 mol) 11.2 L (0.5 mol)
No. of moles of reactant reacted and product formed 0.5 0.5 1
A mount of HCl formed 1 mol

11th Chemistry Chapter 1 Solutions Question 10.
Hot concentrated sulfuric acid is a moderately strong oxidizing agent. Which of the following reactions does not show oxidizing behavior?
(a) Cu + 2H2SO4 → CuSO4 + SO2 + 2H2O
(b) C + 2H2SO4 → 4 CO2 + 2SO2 + 2H2O
(c) BaCl2 + H2SO4 → BaSO4+ 2HCl
(d) none of the above
Answer:
(c) BaCl2 + H2SO4 → BaSO4+ 2HCl
Basic Concepts Of Chemistry And Chemical Calculations Samacheer Kalvi 11th Chemistry Solutions Chapter 1

11th Chemistry Chapter 1 Book Back Answers Question 11.
Choose the disproportional reaction among the following redox reactions.
(a) 3Mg (s) + N2(g) → Mg2N2 (s)
(b) P4 (s) + 3NaOH + 3H2O → PH3(g) + 3NaH2PO2 (aq)
(c) Cl2 (g) + 2Kl (aq) → 2KC1 (aq) + I2
(d) Cr2O3 (s) + 2Al (s) → A2O3 (s) + 2Cr (s)
Answer:
(b) P4 (s) + 3NaOH + 3H2O → PH3(g) + 3NaH2PO2 (aq)
Basic Concepts Of Chemistry And Chemical Calculations Book Back Answers Samacheer Kalvi 11th Chemistry Solutions Chapter 1

Basic Concepts Of Chemistry And Chemical Calculations Pdf Question 12.
The equivalent mass of potassium permanganate in alkaline medium is
MnO4 + 2H2O + 3e → MnO2 + 4OH
(a) 31.6
(b) 52.7
(c) 79
(d) None of these
Answer:
(b) 52.7
The reduction reaction of the oxidizing agent(MnO4) involves gain of 3 electrons.
Hence the equivalent mass = (Molar mass of KMnO4) / 3 = 158.1 / 3 = 52.7

11th Chemistry 1st Chapter Question 13.
Which one of the following represents 180 g of water?
(a) 5 Moles of water
(b) 90 moles of water
(c) \(\frac{6.022 \times 10^{23}}{180}\) Molecules of water
(d) 6.022 × 1024 Molecules of water
Answer:
(d) 6.022 x 1024 Molecules of water
No. of moles of water present in 180 g
= Mass of water / Molar mass of water
= 180 g /18 g mol-1 = 10 moles
One mole of water contains
= 6.022 × 1023 water molecules
10 mole of water contains = 6.022 × 1023 × 10
= 6.022 × 1024 water molecules

Class 11 Chemistry All Formulas Question 14.
7.5 g of a gas occupies a volume of 5.6 liters at 0°C and 1 atm pressure. The gas is …………
(a) NO
(b) N2O
(c) CO
(d) CO2
Answer:
(a) NO
7.5 g of gas occupies a volume of 5.6 liters at 273 K and 1 atm pressure Therefore, the mass of gas that occupies a volume of 22.4 liters –
\(\frac {7.5 g}{5.6 L}\) × 22. 4 L = 30g
Molar mass of NO (14 + 16) = 30g

11th Chemistry Solutions Samacheer Kalvi Question 15.
Total number of electrons present in 1.7 g of ammonia is ………..
(a) 6.022 × 1023
(b) \(\frac{6.022 \times 10^{22}}{1.7}\)
(c) \(\frac{6.022 \times 10^{24}}{1.7}\)
(d) \(\frac{6.022 \times 10^{23}}{1.7}\)
Answer:
(a) 6.022 × 1023
No. of electrons present in one ammonia (NH3) molecule (7 + 3) = 10
No. of moles of ammonia = \(\frac {Mass}{Molar mass}\)
= \(\frac{1.7 \mathrm{g}}{17 \mathrm{g} \mathrm{mol}^{-1}}\) = 0.1 mol
No. of molecules present in 0ne ammonia
= 0.1 × 6.022 × 1023 = 6.O22 × 1022
No. of electrons present in 0.1 mol of ammonia
10× 6.022 × 1022 = 6.022 × 1023

Samacheer Kalvi Class 11 Chemistry Solutions Question 16.
The correct increasing order of the oxidation state of sulphur in the anions SO42-, SO32-, S2O42-,S2O62- is ………..
(a) SO32- < SO32- < S2O42- < S2O62-
(b) SO42- < S2O42- < S2O62-<SO32-
(c) S2O42- < SO32- < S2O62- < SO42-
(d) S2O62- < S2O42- < SO42- < SO32-
Answer:
(c) S2O42- < SO32- < S2O62- < SO42-
Samacheer Kalvi Guru 11th Chemistry Solutions Chapter 1 Basic Concepts Of Chemistry And Chemical Calculations

Class 11 Chemistry Solutions Samacheer Kalvi Question 17.
The equivalent mass of ferrous oxalate is ……….
11th Chemistry Samacheer Kalvi Solutions Chapter 1 Basic Concepts Of Chemistry And Chemical Calculations
Answer:
11th Chemistry Lesson 1 Samacheer Kalvi Basic Concepts Of Chemistry And Chemical Calculations
Chemistry Class 11 Samacheer Kalvi Chapter 1 Basic Concepts Of Chemistry And Chemical Calculations

Samacheer Kalvi 11 Chemistry Solutions Question 18.
If Avogadro number were changed from 6.022 × 1023 to 6.022 × 1020, this would change ………..
(a) the ratio of chemical species to each other in a balanced equation
(b) the ratio of elements to each other in a compound
(c) the definition of mass in units of grams
(d) the mass of one mole of carbon
Answer:
(d) the mass of one mole of carbon

Samacheerkalvi.Guru 11th Chemistry Question 19.
Two 22.4 liter containers A and B contains 8 g of O2 and 8 g of SO2 respectively, at 273 K and 1 atm pressure, then ……….
(a) number of molecules in A and B are same
(b) number of molecules in B is more than that in A
(c) the ratio between the number of molecules in A to the number of molecules in B is 2 : 1
(d) number of molecules in B is three times greater than the number of molecules in A
Answer:
(c) The ratio between the number of molecules in A to number of molecules in B is 2 : 1

Question 20.
What is the mass of precipitate formed when 50 ml of 8.5% solution of Ag NO3 is mixed with 100 ml of 1.865% potassium chloride solution?
(a) 3.59 g
(b) 7 g
(c) 14 g
(d) 28 g
Answer:
(a) 3.59 g
AgNO3 + KCl → KNO3 + AgCl
Solution:
50 mL of 8.5% solution contains 4.25 g of AgNO3
No. of moles of AgNO3 present in 50 mL of 8.5% AgNO3 solution
= Mass / Molar mass = 4.25 / 170 = 0.025 moles
Similarly, No of moles of KCl present in loo mL of 1.865% KCl solution
= 1.865 / 74.5 = 0.025 moles
So total amount of AgCl formed is 0.025 moles (based on the stoichiometry calculator)
Amount of AgCl present in 0.025 moles of AgCl
= no. of moles × molar mass
= 0.025 × 143.5 = 3.59 g

Question 21.
The mass of a gas that occupies a volume of 612.5 ml at room temperature and pressure (25°C and 1 atm pressure) is 1.1g. The molar mass of the gas is ………..
(a) 66.25 g mol-1
(b) 44 g mol-1
(c) 24.5 g mol -1
(d) 662.5 g mol-1
Answer:
(b) 44 g mol-1
Solution:
No. of moles of a gas that occupies a volume of 6 12.5 ml at room temperature and pressure
(25° C and 1 atm pressure)
= 612.5 × 10-3 L/24.5 L mol-1
= 0.02 5 moles
We know that,
Molar mass = Mass / no. of moles
= 1.1 g/0.025 mol = 44 g mol-1

Question 22.
Which of the following contain same number of carbon atoms as in 6 g of carbon -12?
(a) 7.5 g ethane
(b) 8 g methane
(c) both (a) and (b)
(d) none of these
Answer:
(c) both (a) and (b)
Solution:
No. of moles of carbon present in 6 g of C – 12 = Mass / Molar mass
= 6/12 = 0.5 moles = 0.5 × 6.022 × 1023 carbon atoms.
No. of moles in 8 g of methane = 8 116 = 0.5 moles
= 0.5 × 6.022 × 1023 carbon atoms.
No. of moles in 7.5 g of ethane = 7.5 / 16 = 0.25 moles
= 2 × 0.25 × 6.022 × 1023 carbon atoms.

Question 23.
Which of the following compound( s) has/have percentage of carbon same as that in ethylene (C2H4)?
(a) propene
(b) ethyne
(c) benzene
(d) ethane
Answer:
(a) propene
Solution:
Molar mass of carbon
Percentage of carbon in ethylene(C2H6) = Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts Of Chemistry And Chemical Calculations
= \(\frac {24}{28}\) × 100 = 85.71%
Percentage of carbon in propene (C3H6) = \(\frac {24}{28}\) × 100 = 85.7 1%

Question 24.
Which of the following is/are true with respect to carbon – 12?
(a) relative atomic mass is 12 u
(b) oxidation number of carbon is +4 in all its compounds.
(c) I mole of carbon -12 contain 6.022 × 1022 carbon atoms.
(d) all of these
Answer:
(a) relative atomic mass is 12 u

Question 25.
Which one of the following is used as a standard for atomic mass?
(a) 6C12
(b) 7C12
(c) 6C13
(d) 6C14
Answer:
(a) 6C12

II. Write brief answer to the following questions

Question 26.
Define relative atomic mass.

On the basis of carbon, the relative atomic mass of element is defined as the ratio of mass of one atom of the element to the mass of l/12th mass of one atom of Carbon – 12.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts Of Chemistry And Chemical Calculations

Question 27.
What do you understand by the term mole?
Answer:
The mole is defined as the amount of a substance which contains 6.023 x 1023 particles such as atoms, molecules or ions. It is represented by the symbol

Question 28.
Define equivalent mass.
Answer:
The equivalent mass of an element is the number of parts of the mass of an element which combines with or displaces 1.008 parts of hydrogen or 8 parts of oxygen or 35.5 parts of chlorine.

Question 29.
What do you understand by the term oxidation number?
Answer:
Oxidation number refers to the number of charges an atom would have in a molecule or an ionic compound, if electrons were transferred completely. The oxidation numbers reflect the number of electrons “transferred”.

Question 30.
Distinguish between oxidation and reduction.
Answer:
Oxidation:
According to the classical concept, oxidation is a process of addition of oxygen or removal of hydrogen.
Removal of hydrogen
2H2S + O2 → H2O + 2S
Addition of oxygen
C + O2 → CO2
According to the electronic concept, loss of electrons is called oxidation reaction.
Ca → Ca2+ + 2e
During oxidation, oxidation number increases.
Dining oxidation, reducing agent gets oxidised.

Reduction:
Reduction is a process of removal of oxygen or addition of hydrogen.
Addition of hydrogen
Ca + H2 → CaH2
Removal of oxygen
Zn O + C → Zn + CO
According to the electronic concept, gain of electrons is called reduction reaction.
Zn2+ + 2e → Zn
During reduction, oxidation number decreases.
During reduction, oxidising agent gets reduced.

Question 31.
Calculate the molar mass of the following compounds.

  1. urea [CO(NH2)2]
  2. acetone [CH3COCH3]
  3. boric acid [H3BO3]
  4. sulphuric acid [H2SO4]

Answer:
1. urea [CO(NH2)2]
Atomic mass of C =12
Atomic mass of O =16
Atomic mass of 2(N) = 28
Atomic mass of 4(H) = 4
∴ Molar mass of Urea = 60

2. Acetone [CH3COCH3]
Atomic mass of 3(C) = 36
Atomic mass of 1(0) = 16
Atomic mass of 6(H) = 6
∴ Molar mass of Acetone = 58

3. Boric acid [H3BO3]
Atomic mass of B = 10
Atomic mass of 3(H) = 3
Atomic mass of 3(O) = 48
∴ Molar mass of Boric acid = 61

4. Sulphuric acid 2[H2SO4]
Atomic mass of 2(H) = 2
Atomic mass of 1(S) = 32
Atomic mass of 4(O) = 64
∴ Molar mass of Sulphuric acid = 98

Question 32.
The density of carbon dioxide is equal to 1.977 kg m-3 at 273 K and 1 atm pressure. Calculate the molar mass of CO2
Answer:
Molecular mass = Density x Molar volume
Molar volume of CO2 = 2.24 x 10-2 m3
Density of CO2 = 1.977 kg m-3
Molecular mass of CO2 = 1.977 x 103 gm-3  x 2.24 x 10-2 m3
= 1.977 × 10-1 × 2.24 = 44 g

Question 33.
Which contains the greatest number of moles of oxygen atoms?

  1. 1 mol of ethanol
  2. 1 mol of formic acid
  3. 1 mol of H2O

Answer:
1. 1 mol of ethanol
C2H5OH (ethanol) – Molar mass = 24 + 6 + 16 = 46
46 g of ethanol contains 1 × 6.023 × 1023 number of oxygen atoms.

2. 1 mol of formic acid.
HCOOH (formic acid) – Molar mass = 2+12 + 32 = 46
46 g of HCOOH contains 2 × 6.023 × 1023 number of oxygen atoms.

3. 1 mol of H2O
H2O (water) – Molar mass = 2 + 16 = 18
18 g of water contains 1 × 6.023 × 1023 number of oxygen atoms.
∴ 1 mole of formic acid contains the greatest number of oxygen atoms.

Question 34.
Calculate the average atomic mass of naturally occurring magnesium using the following data

Isotope Isotopic atomic mass Abundance (%)
Mg24 23.99 78.99
Mg26 24.99 10
Mg25 25.98 11.01

Answer:
Isotopes of Mg.
Atomic mass = Mg24 = 23.99 x \(\frac {783. 99}{100}\) = 18.95
Atomic mass = Mg26 = 24.99 x \(\frac {10}{100}\) = 2.499
Atomic mass = Mg25 = 25.98 x \(\frac {11.01}{100}\) = 2.860
Average Atomic mass = 24.309
Average atomic mass of Mg = 24.309

Question 35.
In a reaction x + y + z2 → xyz2, identify the limiting reagent if any, in the following reaction mixtures.
(a) 200 atoms of x + 200 atoms of y + 50 molecules of z2
(b) 1 mol of x + 1 mol of y + 3 mol of z2
(c) 50 atoms of x + 25 atoms of y + 50 molecules of z2
(d) 2.5 mol of x + 5 mol of y + 5 mol of z2
Answer:
x + y + z2
(a) 200 atoms of x + 200 atoms of y + 50 molecules of z2 According to the reaction, 1 atom of x reacts with one atom of y and one molecule of z to give product. In the case (a) 200 atoms of x, 200 atoms of y react with 50 molecules of z2 (4 part) i.e. 50 molecules of z2 react with 50 atoms of x and 50 atoms of y. Hence z is the limiting reagent.

(b) 1 mol of x + 1 mol of y + 3 mol of z2
According to the equation 1 mole of z2 only react with one mole of x and one mole of y. If 3 moles of z2 are there, z is limiting reagent.

(c) 50 atoms of x + 25 atoms of y + 50 molecules of z2
25 atoms of y react with 25 atoms of x and 25 molecules of z2. So y is the limiting reagent.

(d) 2.5 mol of x + 5 mol of y + 5 mol of z2
2.5 mol of x react with 2.5 mole of y and 2.5 mole of z2. So x is the limiting reagent.

Question 36.
Mass of one atom of an element is 6.645 × 10-23 g. How many moles of element are there in 0.320 kg?
Answer:
Mass of one atom of an element = 6.645 × 10-23 g = Atomic mass.
Mass of given element = 0.320 kg
Number of moles =
Atomic mass
11th Chemistry Chapter 1 Solutions Basic Concepts Of Chemistry And Chemical Calculations
= 48.156 x 10-23
= 4.8156 x 10-24 moles.

Question 37.
What is the difference between molecular mass and molar mass? Calculate the molecular mass and molar mass for carbon monoxide.
Answer:
Molecular mass:

  • Relative molecular mass is defined as the ratio of the mass of the molecule to the unified atomic mass unit.
  •  It can be calculated by adding the relative atomic masses of its constituent atoms.
  • For carbon monoxide (CO) Molecular mass = Atomic mass of carbon + Atomic mass of oxygen 12 + 16 = 28 u.

Molar mass:

  • It is defined as the mass of one mole of a substance.
  • The molar mass of a compound is equal to the sum of the relative atomic masses of its constituent expressed in g mol-1.
  •  For carbon monoxide (CO) 12 + 16 = 28 g mol-1 Both molecular mass and molar mass are numerically same but the units are different.

Question 38.
What is the empirical formula of the following?

  1. Fructose (C6H12O6) found in honey
  2. Caffeine (C8H10N4O2) a substance found in tea and coffee.

Answer:
1. Fructose (C6H12O6)
Empirical formula is the simplest formula. So it is divided by 6 and so empirical formula is CH2O.

2. Caffeine (C8H10N4O2)
Simplified formula = \(\frac {molecular formula}{2}\)
Empirical formula = C4H5N2O.

Question 39.
The reaction between aluminium and ferric oxide can generate temperatures up to 3273 K and is used in welding metals. (Atomic mass of AC = 21 u Atomic mass of 0 = 16 u) 2Al + Fe2O2 → Al2O3 + 2Fe; If, in this process, 324 g of aluminium is allowed to react with 1.12 kg of ferric oxide.

  1. Calculate the mass of Al2O3 formed.
  2. How much of the excess reagent is left at the end of the reaction?

Answer:
11th Chemistry Chapter 1 Book Back Answers Basic Concepts Of Chemistry And Chemical Calculations
1. As per balanced equation 54 g A1 is required for 112 g of iron and 102 g of Al2O3.
54 g of Al gives 102 g of Al2O3.
∴ 324 g of Al will give \(\frac{102}{54}\) x 324 = 612 g of Al2O3.

2. 54 g of Al requires 160 g of Fe2O3 for welding reaction.
∴ 324 g of Al will require \(\frac {160}{54}\) x 324 = 960 g of Fe2O3.
∴ Excess Fe2O3 – Un reacted Fe2O3 = 1120 – 960 = 160 g
160 g of excess reagent is left at the end of the reaction.

Question 40.
How many moles of ethane is required to produce 44 g of CO2 (g) after combustion.
Answer:
Basic Concepts Of Chemistry And Chemical Calculations Pdf Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic
∴ 44g of CO2 = I mole of CO2
2 moles of CO2 is produced by 1 mole of ethane.
∴ 1 mole of CO2 will be produced by = ?
∴ To produce 1 mole of CO2, the required mole of ethane is = \(\frac {1}{2}\) x 1 = 0.5 mole of ethane.

Question 41.
Hydrogen peroxide is an oxidizing agent. It dioxides ferrous ion to ferric ion and reduced itself to water. Write a balanced equation.
Answer:
H2O2 – Oxidizing agent
Fe2+ + H2O2 → Fe3+ + H2O (Acetic medium)
Ferrous ion is oxidized by H2O2 to Ferric ion.
The balanced equation is Fe2+ → Fe3+ + e x 2
11th Chemistry 1st Chapter Samacheer Kalvi Basic Concepts Of Chemistry And Chemical Calculations

Question 42.
Calculate the empirical and molecular formula of a compound containing 76.6% carbon, 6.38 % hydrogen and rest oxygen its vapour density is 47.
Answer:
Class 11 Chemistry All Formulas Samacheer Kalvi Chapter 1 Basic Concepts Of Chemistry And Chemical Calculations
Empirical formula = C6H6O
Va-pour density 47
∴ Molecular mass = 2 x vapor density = 2 x 47 = 94
Molecular formula Empirical formula x n
Molecular mass x n
n = \(\frac { Molecular mass }{ Empirical formula mass }\) = \(\frac {94}{94}\) = 1
∴ Molecular formula = C6H6O

Question 43.
A Compound on analysis gave Na = 14.31% S = 9.97% H = 6.22% and O = 69.5% calculate the molecular formula of the compound if all the hydrogen in the compound is present in combination with oxygen as water of crystallization, (molecular mass of the compound is 322).
Answer:
11th Chemistry Solutions Samacheer Kalvi Chapter 1 Basic Concepts Of Chemistry And Chemical Calculations
All H combines with 10 oxygen atoms to form as 10H2O.
So the empirical formula is Na2SO4 .10H2 O
Empirical formula mass = (23 x 2) + (32 x 1) + (16 x 4) + (10 x 18)
= 46 + 32 + 64 + 180 = 322
n = \(\frac { Molecular mass }{ Empirical formula mass }\) = \(\frac {322}{322}\) = 1
Molecular formula = Na2SO4. 10H2O

Question 44.
Balance the following equations by oxidation number method

  1. K2Cr2 O7 + KI + H2SO2 → K2SO4 + Cr2(SO4)3 + I2 + H2O
  2. KMnO4 + Na2SO3 → Mn02 + Na2SO4 + KOH
  3. Cu+ HNO3 → Cu(N03)2 + NO2 + H2O
  4. H2C2O4 + KMnO4 + H2SO4 → K2SO4 + MnSO4 + CO2 + H2O

Answer:
1. K2Cr2 O7 + KI + H2SO2 → K2SO2 + Cr2(SO4)3 + I2 + H2O
Step – 1.
Samacheer Kalvi Class 11 Chemistry Solutions Chapter 1 Basic Concepts Of Chemistry And Chemical Calculations
Step – 2
K2Cr2 O7 + 6KI + H2SO4 → K2SO4 + Cr2(SO4)3 + 3I2 + H2O
Step – 3
To balance other atoms
K2Cr2 O7 + 6KI + H2SO4 → 4K2SO4 + Cr2(SO4)3 + 3I2 + H2O
Step – 4
K2Cr2 O7 + 6KI + 7H2SO4 → 4K2SO4 + Cr2(SO4)3 + 3I2 + 7H2O

2. KMnO4 + Na2SO3 → MnO2 + Na2SO4 + KOH (Alkaline medium)
Step – 1
Class 11 Chemistry Solutions Samacheer Kalvi Chapter 1 Basic Concepts Of Chemistry And Chemical Calculations
Step – 2
2KMnO4 + 3Na2SO3 → 2MnO2 + 3Na2SO4 + KOH
Step – 3
balancing potassium, KOH is multiplied by 2
2KMnO4 + 3Na2SO3 → 2MnO2 + 3Na2SO4 + KOH
Step – 4
To balance H atom, H20 is added on reactant side.
2KMnO4 + 3Na2SO3 + H2O → 2MnO2 + 3Na2SO4 + KOH

3. Cu + HNO3 → Cu(NO3)2 + NO2 + H2O
Step – 1
Samacheer Kalvi 11 Chemistry Solutions Chapter 1 Basic Concepts Of Chemistry And Chemical Calculations
Step – 2
Cu + HNO3 → Cu(NO3)2 + NO2 + H2O
Step – 3
To balance Nitrogen, 2HNO3 is multiplied by 2 and NO2 is multiplied by 2
Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + H2O
Step 4.
To balance oxygen, H2O is multiplied by 2
Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + 2H2O

4. H2C2O4 + KMnO4 + H2SO4 → K2SO4 + MnSO4 + CO2 + H2O
Step – 1
Samacheerkalvi.Guru 11th Chemistry Solutions Chapter 1 Basic Concepts Of Chemistry And Chemical Calculations
Step – 2
5 H2C2O4 + KMnO4 + H2SO4 → K2SO4 + MnSO4 + 10 CO2 + H2O
Step – 3
To balance K, KMnO4 and MnSO4 are multiplied by 2
5 H2C2O4 + 2KMnO4 + H2SO4 → K2SO4 + 2MnSO4 + 10 CO2 + H2O
Step – 4
To balance O and H, H2O and H2SO4 are multiplied by 3 and 6.
5 H2C2O4 + 2KMnO4 + 3H2SO4 → K2SO4 + 2MnSO4 + 10 CO2 + 8H2O

Question 45.
Balance the following equations by ion electron method.

  1. KMnO4 + SnCl2  + HCl → MnCl2  + SnCl4  + H2O + KCl
  2. C2O42- + Cr2 O7 2- → Cr 3+ + CO2  (in acid medium)
  3. Na2S2O3 + I2 → Na2S4O6 + NaI (in acid medium)
  4. Zn + NO3 → Zn2+ + NO

Answer:
1. KMnO4 + SnCl2 + HCl → MnCl3 + SnCl3 +H2O + KCl
Oxidation half reaction: (loss of electrons)
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Reduction half reaction: (gain of electrons)
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Add H2O to balance oxygen atoms.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Add HCl to balance hydrogen atoms
KMnO4 + 5e + 8HCl → MnCl2 + 4H2O ………(4)
To equalize the number of electrons equation (1) x 5 and equation (2) x 2
5SnCl2 → 5SnCl4 + 10e
2KMnO4 + 16HCl + 10e → 2MnCl2 + 4H2O + 2KCl
2KMnO4 + 5SnCl2 + 16HCl → 5SnCl4 + 2MnCl2 + 4H2O + 2KCl

2. C2O42- + Cr2 O72- → Cr 3+ + CO2  (in acid medium)
Oxidation half reaction:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Reduction half reaction:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
To balance oxygen atoms, H2O is added on RHS of equation (2)
Cr2O72- + 6e → 2Cr3+ + 7 H2O ……….(3)
To balance Hydrogen atoms, H+ is added on LHS of equation (1)
C2O42- + 14H+ → 2CO2 + 2e ……..(4)
To equalize the number of electrons gained and lost, multiply the equation (4) x 3.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

3. Na2S2O3 + I2 → Na2S4O6 + NaI (in acid medium)
Oxidation half reaction: (Loss of electron)
Na2S2O3 → Na2S4O6 + 2e2- ……..(1)
Reduction half reaction: (Gain of electron)
I2 + 2e2-→ 2NaI …………(2)
Adding (1) and (2)
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
To balance oxygen,
2Na2S2O3 + I2 → Na2S2O2 + 2NaI

In acidic medium
4. Zn + NO3 → Zn2+ + NO
Half reactions are –
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Samacheer Kalvi 11th Chemistry Basic Concepts of Chemistry and Chemical Calculations In Text Questions – Evaluate Yourself

Question 1.
By applying the knowledge of chemical classification, classify each of the following Into elements, compounds or mixtures.

  1. Sugar
  2. Sea water
  3. Distilled water
  4. Carbon dioxide
  5. Copper wire
  6. Table salt
  7. Silver plate
  8. Naphthalene balls

Answer:

Elements Compounds Mixtures
Copper wire (Cu)
Silver plate (Ag)
Sugar
Distilled water Carbon dioxide Naphthalene balls
Sea water Table salt

Question 2.
Calculate the molar mass of the following.

  1. Ethanol (C2H5OH)
  2. Potassium permanganate (KMnO4)
  3. Potassium dichromate (K2Cr2O7)
  4. Sucrose (C12H22O11)

Answer:
(1) Ethanol (C2H5OH)
Molar mass = (2 × 12) + (6 × 1) + (1 × 16)
= 24 + 6+16 = 46 g mol-1

(2) Potassium permanganate (KMnO4)
Molar mass = 39 + 55 + (4 × 16)
= 39 + 55 + 64 = 158 g mol-1

(3) Potassium dichromate (K2Cr2O7)
Molar mass = (39 × 2) + (2 × 52) + (7 × 16)
= 78 + 104 + 112 = 294 g mol-1

(4) Sucrose (C12H22O11)
Molar mass = (12 × 12) + (22 × 1) + (11 × 16) = 144 + 22 + 176 = 342 g mol-1

Question 3.
(a) Calculate the number of moles present in 9 g of ethane.
Answer:
Mass of ethane = 9 g
Molar mass of ethane C2H6 = 30 g mol-1
No. of moles = \(\frac {Mass}{Molar mass}\) = \(\frac {9}{30}\) = 0.3 mol.

(b) Calculate the number of molecules of oxygen gas that occupies a volume of 224 ml at 273 K and 3 atm pressure.
Answer:
Molar volume of oxygen = 22400 ml.
22400 ml of oxygen contains 6.023 x 1023 molecules.
224 ml of oxygen contain \(\frac{6.023 \times 10^{23}}{22400}\) x 224
\(\frac{6.023 \times 10^{23}}{100}\) = 6.023 × 1021

Question 4.
(a) 0.456 g of a metal gives 0.606 g of its chloride. Calculate the equivalent mass of the metal.
Answer:
Mass of the metal = W1 = 0.456 g.
Mass of the metal chloride = W2 = 0.606 g.
∴Mass of chlorine = W2 – W1 = 0.606 – 0.456 = 0.15 g
0.15 g of chlorine combine with 0.456 g of metal.
∴ 35.46 g of chlorine will combine with \(\frac {0.456}{0.15}\) x 35.46 = 107.79 g eq-1

(b) Calculate the equivalent mass of potassium dichromate. The reduction half – reaction in acid medium is –
Cr2O72- + 14H+ +6e → 2 Cr3+ + 7H2O
Answer:
Cr2O72- + 14H+ +6e → 2 Cr3+ + 7H2O
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Molar mass of K2Cr2O7 = 294.18
∴ Equivalent mass of K2Cr2O7 = \(\frac { 294.18}{6}\) = 49.03

Question 5.
A Compound on analysis gave the following percentage composition C = 54.55%, H = 9.09%, O = 36.36%. Determine the empirical formula of the compound.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Empirical formula = C2H4O

Question 6.
Experimental analysis of a compound containing the elements x,y,z on analysis gave the following data, x = 32 %, y = 24 %, z = 44 %. The relative number of atoms of x, y and z are 2,1 and 0.5, respectively. (Molecular mass of the compound is 400 g) Find out.

  1. The atomic masses of the element x,y,z.
  2. Empirical formula of the compound and
  3. Molecular formula of the compound.

Answer:
Element x = 32% ; y = 24% ; z = 44%
Relative number of atoms x = 2 ; y = 1 ; z = 0.5
Molar mass of the compound = 400 g.

1. Atomic mass of the element.
Relative number of atoms Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
∴  Atomic mass Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Atomic mass x = \(\frac {32}{2}\) = 16
Atomic mass y = \(\frac {24}{1}\) = 24
Atomic mass z = \(\frac {44}{0.5}\) = 88
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

2. Empirical formula of the compound x4 y2 Z1
Molecular mass of the compound = 400
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations = \(\frac {400}{200}\) = 2

3. Molecular formula of the compound = x8 y4 z2

Question 7.
The balanced equation for a reaction is given below 2x + 3y → 41 + m When 8 moles of x react with 15 moles of y, then –

  1. Which is the limiting reagent?
  2. Calculate the amount of products formed.
  3. Calculate the amount of excess reactant left at the end of the reaction.

Answer:
2 x + 3 y → 41 + m

1. 2x reacts with 3y to give products.
5x reacts with l5y means, y is the excess because 8 moles of x should react withn 4 x 3y = 12y moles of y to give products. In this reaction 15y moles are used. Therefore, 3 moles of y is excess and it is the limiting agent.

2. When 8 moles of x react with 12 moles of y, the product formed will be 4 x 41 i.e. 161 and 4m as product.
8x + 12y → 161 + 4m

3.  At the end of the reaction, the excess reactant left in 3 moles of y.

Question 8.
Balance the following equation using oxidation number method
AS2 S3  + HNO3 + H2O → H3ASO4 + H2SO4 + NO
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
To balance oxygen and sulphur, H2O and H2SO4 are added.
3AS2S3 + 2HNO3 + H2O → 6H3ASO4 + 2NO + H2SO2
3AS2S3 + 28HNO3 + 4H2O → 6H3ASO4 + 28NO + 9H2SO4

Samacheer Kalvi 11th Chemistry Solutions Basic Concepts of Chemistry and Chemical Calculations Textual Calculations based on Stolchlometry Solved

Question 1.
How many moles of hydrogen is required to produce 10 moles of ammonia?
Answer:
The balanced stoichiometric equation fòr the formation of ammonia is
N2(g) + 3H2(g) → 2NH2(g)
4s per the stoichiometric equation, to produce 2 moles of ammonia, 3 moles of hydrogen are required.
∴ to produce 10 moles of ammonia,
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
= 15 moles of hydrogen are required.

Question 2.
Calculate the amount of water produced by the combustion of 32 g of methane.
Answer:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
As per the stoichiometric equation,
Combustion of 1 mole (16 g) CH4 produces 2 moles (2 x 18 = 36 g) of water.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Combustion of 32 g CH4 produces
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 3.
How much volume of carbon dioxide is produced when 50 g of calcium carbonate is heated completely under standard conditions?
Answer:
The balanced chemical equation is,
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
As per the stoichiometric equation,
1 mole (100g) CaCO3 on heating produces 1 mole CO2
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
At STP, 1 mole of CO2 occupies a volume of 22.7 litres
At STP, 50 g of CaCO3 on heating produces,
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
= 11.35 litres of CO2

Question 4.
How much volume of chlorine is required to form 11.2 L of HCl at 273 K and 1 atm pressure ?
Answer:
The balanced equation for the formation of HCl is,
H2 (g) + Cl2 (g) → 2 HCl (g)
As per the stoichiometric equation, under given conditions,
To produce 2 moles of HCl, 1 mole of chlorine gas is required.
To produce 44.8 liters of HCl, 22.4 liters of chlorine gas are required
∴ To produce 11.2 liters of HCl,
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
= 5.6 litres of chlorine are required.

Question 5.
Calculate the percentage composition of the elements present in magnesium carbonate. How many kilogram of CO2 can be obtained by heating 1 kg of 90 % pure magnesium carbonate.
Answer:
The balanced chemical equation is
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Molar mass of MgCO3 is 84 g mol-1.
84 g MgCO3 contain 24 g of Magnesium.
∴ 100 MgCO3 contain
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
= 28.57 g Mg i.e. percentage of magnesium = 28.57 %.
84 g MgCO3 contain 12 g of carbon
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
= 14.29 g of carbon
Percentage of carbon = 14.29 %.
84 g MgCO3 contain 48 g of oxygen
∴ 100 g MgCO3 contains
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
= 57.14 g of oxygen.
∴ Percentage of oxygen = 57.14 %.
As per the stoichiometric equation,
84 g of 100 % pure Mg CO3 on heating gives 44 g of CO2.
∴1000 g of 90 % pure Mg CO3 gives
\(\frac {44 g}{84 g x 100 %}\) x 90 % x 1000 g
= 471.43 g of CO2
= 0. 471 kg of CO2

Question 6.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
(1) If the entire quantity of all the reactants is not consumed in the reaction which is the limiting reagent ?
(2) Calculate the quantity of urea formed and un reacted quantity of the excess reagent. The balanced equation is
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Answer:
(1) The entire quantity of ammonia is consumed in the reaction. So ammonia is the limiting reagent. Some quantity of CO2 remains unreacted, so CO2 is the excess reagent.

(2) Quantity of urea formed = number of moles of urea formed x molar mass of urea
= 19 moles x 60 g mol-1
= 1140 g = 1.14 kg
Excess reagent leftover at the end of the reaction is carbon dioxide. Amount of carbon dioxide leftover = number of moles of CO2 left over x molar mass of CO2
= 7 moles x 44 g mol-1 = 308 g.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Samacheer Kalvi 11th Chemistry Solutions Basic Concepts of Chemistry and Chemical Calculations Additional Questions Solved

I. Choose the correct answer from the following

Question 1.
Which one of the following is the standard for atomic mass?
(a) 1H1
(b) 66C12
(c) 6C14
(d) 8O16
Answer:
(b) 66C12
Hint:
Standard element used to determine atomic mass is 6 Cl2

Question 2.
One mole of CO2 contains ………….
(a) 6.023 x 1023 atoms of C
(b) 6.023 x 1023 atoms of O
(c) 18.1 x 1023 molecules of CO2
(d) 3g atoms of CO2
Answer:
(a) 6.023 x 1023 atoms of C
Hint:
One mole of any substance contains Avogadro number of atoms. In this carbon one mole is present and oxygen two atoms are present. So, 6.023 x 1023 atoms of C is correct.

Question 3.
The largest number of molecules is in
(a) 54 g of nitrogen pent oxide
(b) 28 g of carbon dioxide
(c) 36 g of water
(d) 46 g of ethyl alcohol
Answer:
(c) 36 g of water
Hint:(a) 54 g of N2O5
N2O5 = Molecular mass = 28 + 80 = 108
108 g of N2O5 contains 6.023 x 1023 molecules.
∴ 54 g of N2O5 will contain \(\frac{6.023 \times 10^{23}}{108}\) x 54 = 3.0115 x 1023 molecules.

(b) 28 g of CO2
CO2 = Molecular mass = 12 + 32 = 44
44 g of CO2 contains 6.023 x 1023 molecules.
∴ 28 g of CO2 will contain \(\frac{6.023 \times 10^{23}}{44}\) x 28 = 3.832 x 1023 molecules.

(c) 36 g of H2O
H2O = Molecular mass = 2 + 16 = 18
18 g of H2O contains 6.023 x 1023 molecules.
∴ 36 g of H2O will contain \(\frac{6.023 \times 10^{23}}{18}\) x 36 = 12.046 X 1023molecules.

(d) 46 g of C2H5OH
C2H5OH = Molecular mass = 24 + 6 + 16 = 46
46 g of C2H5OH contains 6.023 x 1023 molecules.
So, among the 4, 36 g of water contain the largest number of molecules as 12.046 x 1023.

Question 4.
The number of moles of H2 in 0.224 liter of hydrogen gas at STP is …………..
(a) 1
(b) 0.1
(c) 0.01
(d) 0.001
Answer:
(c) 0.01
22.4 liter of hydrogen gas at STP contains 1 mole.
∴ 0.224 liter of hydrogen gas at STP will contain \(\frac{1}{22.4}\) x 0.224 = 0.01

Question 5.
10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. The amount of water produced in this reaction will be ………..
(a) 3 mol
(b) 4 mol
(c) 1 mol
(d) 2 mol
Answer:
(b) 4 mol
10 g of H2 + 64 g of O2 → water
2H2 + O2 → 2H2O
4 g of hydrogen react with 32 g of oxygen. So 10 g of hydrogen will react with 80 g of oxygen. But we have given amount of oxygen only 64 g. It means, here oxygen is the limiting agent. Now all the oxygen react with 8 g of hydrogen and form 4 moles of water.

Question 6.
6.023 x 1020 molecules of urea are present in 100 ml of its solution. The concentration of the solution is –
(a) 0.02 M
(b) 0.1 M
(c) 0.01 M
(d) 0.001 M
Answer:
(c) 0.01 M
100 ml of solution contain 6.023 x 1020 molecules.
6.023 x 1023 molecules in 1000 ml = 1 M.
∴ 6.023 x 1020 molecules in 1000 ml =  Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
= 1020 x 10-23 = 10-3
10-3 moles present in 100 ml.
∴ In 1000 ml the moles present is = \(\frac{10^{-3}}{100}\) x 1000 = 10-2 moles
Concentration = 0.01 M

Question 7.
Two containers A and 13 of equal volume contain 6 g of O2 and SO2 at 300 K and 1 atm. Then
(a) No. of molecules in A is less than that in B
(b) No. of molecules in A is more than that in B
(c) No. of molecules in A and B are same
(d) none of these
Answer:
(b) No, of molecules in A is more than that in B
O2 = Mass = 6 g
Molar mass = 32 g
32 g of O2 contains 6.023 x 1023 molecules.
∴ 6 g of O2 will contain = 6.023 X 1023 x 6 = 1.129 x 1023 molecules.
SO2 = Mass = 6 g
Molar mass = 64 g
64 g of SO2 contains 6.023 x 1023 molecules.
∴ 6 g of SO2 will contain 6.023 x 1023 x 6 = 0.5646 x 1023 molecules.
∴ Number of molecules in A is more than that in B.

Question 8.
The number of molecules in 16 g of methane is ……….
(a) 3.023 x 1023
(b) 6.023 x 1023
(c) 16 / 6.023 x 1023
(d) 6.023 / 3 x 1023
Answer:
(b) 6.023 x 1023
Hint:
Methane: CH4
Molecular mass 12 + 4 = 16
16 g of methane contains Avogadro number of molecules 6.023 x 1023 molecules.

Question 9.
Number of atoms in 4.25 g of ammonia is …………..
(a) 1 x 1023
(b) 2 x 1023
(e) 4 x 1023
(d) 6 x 1023
Answer:
(d) 6 x 1023
Hint:
Ammonia = NH3 (4 atoms)
Molecular mass = 14 + 3 = 17
17 g of Ammonia contains 6.023 X 1023 atoms.
∴ 4.25 g of Ammonia will contain 6.0231; 1023 x 4.25 = 1.5055 x 1023 molecules.
∴ 1.5055 x 1023 molecules will contain 4 x 1.5 x 1023 = 6 x 1023 molecules.

Question 10.
The number of molecules in a drop of water (0.0018 ml) at room temperature is ……….
(a) 6.02 x 1023
(b) 1.084 x 1023
(c) 4.84 x 1023
(d) 6.02 x 1023
Answer:
(a) 6.02 x 1023
Hint:
0.0018 ml – drop of water = 0.0018 g
H2O = molecular mass = 18 g.
Number of molecules in 18 g = 6.023 x 1023
∴ Number of molecules in 0.0018 g 6.023 x 1023 x 0.0018 = 6.023 x 1023 x 105
= 6.023 x l019 molecules.
or
Density of water at 25°C = 997.0479 g / L
Mass of 0.0018 ml (or) 0.00 18 x 10-3 L
= D x V
= 997.05 x 0.0018 x 10-3
= 1.795 x 10-3 g
Molar mass of water = 18 g
Mole = \(\frac{Mass}{Molecular mass}\) = \(\frac{1.795 \times 10^{-3}}{18}\)
= 9.971 x 10-5
∴ Number of molecules in 0.0018 ml = moles x Avogadro number
= 9.971 x 10-5 x 6.023 x 1023
= 6 x 1019 molecules.

Question 11.
7.5 g of a gas occupy 5.6 liters of volume at STP. The gas ……….
(a) NO
(b) N2O
(c) CO
(d) CO2
Answer:
(a) NO
Hint:
22.4 liters = 1 mole
5.6 liters = \(\frac {1}{22.4}\) x 5.6 = 0.25 mole.
NO = molar mass = 14 + 16 = 30 g = 1 mole
N2O = molar mass = 28 + 16 = 44g = 1 mole
CO = molar mass = 12 + 16 = 28 g = 1 mole
CO2=molar mass = 12 + 32 = 44g = 1 mole
Among the four gases, 0.25 mole = 7.5 g is equal to NO gas.

Question 12.
The mass in grams of 0.45 mole of CO2 ions ……….
(a) 1.8
(b) 40
(c) 36
(d) 18
Answer:
(d) 18
Hint:
Ca = Atomic mass = 40
Ca → Ca2+ + 2e
41 g of Ca = 1 mole (for Ca2+ Atomic mass remains same)
1 mole of Ca2+ = 40 g
∴ 0.45 mole of Ca2+ = \(\frac {40}{1}\) x 0.45 = 18g

Question 13.
The mass of one molecule of HI in grams is ……….
(a) 2.125 x 10-22
(b) 128
(c) 127
(d) 6.02 x 10-23
Answer:
(b) 128
Hint:
HI = 1 mole = 1 + 127 = 128 g

Question 14.
Avogadro’s number is the number of molecules present in ……….
(a) 1 g of molecule
(b) 1 g atom of molecule
(c) gram molecular mass
(d) I lit of molecule
Answer:
(c) gram molecular mass
by definition (c) is correct

Question 15.
Which of the following contains same number of carbon atoms as are in 6.0 g of carbon (C-12)?
(a) 6.0 g ethane
(b) 8.0 g methane
(c) 21.0 g Propane
(d) 28.0 g CO
Answer:
(b) 8.0 g methane

Hint:
(a) 6.0 g of ethane (C2H6)
C2H6 = molar mass = 24 + 6 = 30 g
30 g of ethane contains 2 x 6.023 x 1023 Carbon atoms.

(b) 8.0 g of methane (CH4)
CH4 molar mass = 12 + 4 = 16g
16 g of methane contains 6.023 x 1023 Carbon atoms.

(c) 21.0 g of propane (C3H8)
C3H8 = molar mass = 36 + 8 = 44 g
44 g of propane contains 3 x 6.023 x 1023 Carbon atoms.

(d) 28.0 g of Carbon monoxide (CO)
CO = molar mass = 12+ 16 = 28 g
28 g of Carbon monoxide contains 6.023 x 1023 Carbon atoms.
6.0 g of Carbon contains = 6.023 x 10 x 6 = 3.0115 x 1023 Carbon atoms.
Among the (a), (b), (c), (d) – 8 g of CH4 contains x 8 = 3.0115 x 1023 Carbon atoms.

Question 16.
Equivalent mass of KMnO4 when it is converted to MnSO4 is equal to molar mass divided by ………..
(a) 6
(b) 4
(c) 5
(d) 2
Answer:
(c) 5
Hint:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 17.
How many equivalents of Sodium sulphate is formed when Sulphuric acid is completelyn neutralized with a base NaOH?
(a) 0.2
(b) 2
(e) 0.1
(d) 1
Answer:
(d) 1
Hint:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 18.
Cl2 changes to Cl and ClO in cold NaOH. Equivalent mass of Cl2 will be ………..
(a) Molar mass / 2
(b) Molar mass / 1
(c) Molar mass / 3
(d) 2 x Molar mass / 2
Answer:
(a) Molar mass / 2
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 19.
Equivalent mass of KMnO4 in acidic medium, concentrated alkaline medium and dilute basic medium respectively are M, M, M. Reduced products can be ……………
(a) MnO2, MnO22-, Mn2+
(b) MnO2, Mn2+, MnO42-
(c) Mn2+, MnO2, MnO42-
(d) Mn2+, MnO42-, MnO2
Answer:
(c) Mn2+, MnO2, MnO42-
Hint:
MnO4 + 8H+ + 5e → Mn2+ + 4H2O (acidic medium)
MnO4 + 4H+ + 3e → MnO2 + 2H2O (concentrated basic medium)
MnO4 + e → MnO42- (dilute basic medium)

Question 20.
The empirical formula of hydrogen peroxide is
(a) HO
(b) H2O
(e) H3O
(d) H2O2
Answer:
(a) HO
Hint:
Molecular formula of hydrogen peroxide = H2O2
H2O2 ÷ 2 = HO = Empirical formula.

Question 21.
Molecular mass =
(a) Vapour Density × 2
(b) Vapour Density ÷ 2
(c) Vapour Density × 3
(d) Vapour Density
Answer:
(a) Vapour Density × 2

Question 22.
20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesium oxide. What will be the percentage of purity of magnesium carbonate in the sample?
(a) 60
(b) 84
(e) 75
(d) 96
Answer:
(b) 84
Hint:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
84 g of MgCO3 gives 40 g of MgO. (100% purity)
20 g of MgCO3 will give = \(\frac{40 \times 20}{84}\) = 9.52 g
9.52 g of MgO is given by 100% pure MgCO3.
8.0 g of MgO will be given by = \(\frac{100 \times 8}{9.52}\) = 84.03%

Question 23.
What is the mass of the precipitate formed when the preparation of alkyl halides 50 ml of 16.9 % solution of AgNO3 is mixed with 50 ml of 5.8 % NaCl solution?
(a) 7 g
(b) 14 g
(c) 28 g
(d) 35 g
Answer:
(a) 7 g
Hint:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 24.
When 22 L of hydrogen gas is mixed with 11.2 L of chlorine gas, each at STP, the moles of HCl gas formed is equal to ……….
(a) 2
(b) 0.5
(c) 1.5
(d) 1
Answer:
(d) 1
Hint:
H2 + Cl2 → 2HCl
22 liters + 11.2 litres
1 mole + \(\frac {1}{2}\) mole = 1 mole of HCl and \(\frac {1}{2}\) mole of H2 is remained.

Question 25.
5.6 L of a gas at STP are found to have mass of 11 g. The molecular mass of the gas is
(a) 36
(b) 48
(c) 40
(d) 44
Answer:
(d)44
Hint:
5.6 L of gas has the mass of 11 g.
∴ 22.4 L of gas will have the mass \(\frac {11}{5.6}\) x 22.4 = 44

Question 26.
Oxidation number f Fluorine in all compounds is ……….
(a) + 1
(b) -1
(c) 0
(d) – 2
Answer:
(b) – 1

Question 27.
In redox reaction which of the following is true?
(a) Number of electrons lost is more than number of electrons gained
(b) Number of electrons lost is less than number of electrons gained
(c) Number of electrons lost s equal number of electrons gained
(d) No transfer and gain of electrons during the reaction.
Answer:
(c) Number of electrons lost is equal number of electrons gained

Question 28.
Which of the following is a mono-atomic molecule?
(a) Hydrogen
(b) Oxygen
(c) Sodium
(d) Ozone
Answer:
(c) Sodium

Question 29.
Which one of the following is a diatomic molecule?
(a) Ozone
(b) Copper
(c) Hydrogen
(d) Gold
Answer:
(c) Hydrogen

Question 30.
The value of Avogadro Number N is equal to ……….
(a) 2.24 x 10-2L
(b) 22400 cm3
(c) 6.023 x 10-23
(d) 6.023 x 1023
Answer:
(d) 6.023 x 1023

Question 31.
46 g of ethanol contains ……….
(a) 2 x 6.023 x 1023 C atoms
(b) 3 x 6.023 x 1023 atoms
(c) 9 x 6.023 x 1023 H atoms
(d) 6.023 x 1023 Carbon atoms
Answer:
(a) 2 x 6.023 x 1023 C atoms
Hint:
C2H5OH = Molecular mass = (12 x 2) + (1 x 6) + (1 x 16)
=24 + 6 + 16 = 46
2 Carbon atoms are present.
∴ 2 x 6.023 x 1023 C atoms is correct.

Question 32.
The mass of one mole of CaCl2 is ……….
(a) 55.5 g mol-1
(b) 111 g mol-1
(c) 222 g mol -1
(d) 77.5 g mol-1
Answer:
(b) 111 g mol-1
Hint:
CaCl2 = Mass = 40 + 71 = 111 g mol-1

Question 33.
22 g of CO2 contains molecules of CO2
(a) 6.023 x 1023
(b) 6.023 x 1023
(c) 3.0115 x 1023
(d) 3.0115 x 1023
Answer:
(c) 3.0115 x 1023
Hint:
44 g of CO2 contains 6.023 x 1023 molecules.
∴ 22 g of CO2 will contain = \(\frac{6.023 \times 10^{23}}{44}\) x 22 = 3.0115 x 1023

Question 34.
The formula weight of ethanol (C2H5OH) is ……….
(a) 56.5 amu
(b) 16 amu
(c) 60 amu
(d) 46 amu
Answer:
(d) 46 amu
Hint:
Formula weight of C2H5OH = (12 x 2) + (6 x 1) + (1 x 16)
= 24 + 6 + 16 = 46 amu

Question 35.
The number of moles of ethane in 60 g is ……….
(a) 2
(b) 4
(c) 0.5
(d) 1
Answer:
(a) 2
Hint:
C2H6 – Ethane – molar mass = 24 + 6 = 30 g
30 g of C2H6 contains = 1 mole.
∴ 60 g of C2H6 will contains = x 1 = 2 moles.

Question 36.
Which of the following method is used to prevent rusting of iron?
(a) Galvanization
(b) Painting
(c) Chrome plating
(d) all the above
Answer:
(d) all the above

Question 37.
Which of the following is not a redox reaction?
(a) H2 + F2 → 2HF
(b) Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + 2H2O
(c) 2H2 + O2 → 2H2O
(d) AgCl + NH3 → [Ag(NH3)2]Cl
Answer:
(d) AgCl + NH3 → [Ag(NH3)2]Cl

Question 38.
How many H2O molecules are there in a snowflake weighing 1 mg?
(a) 3.35 x 1019
(b) 6.023 x 1023
(c) 335 x 10-19
(d) 100
Answer:
(a) 3.35 x 1019
Hint:
1 mg of H2O
Molar mass of H2O = 2 + 16 = 18 g
1 Mole contains 6.023 x 1023 water molecules
18 g contains 6.023 x 1023 water molecules
l mg contains \(\frac{6.023 \times 10^{23}}{18}\) x 1g /1000 mg = 3.35 x 10-19 H2O molecule.

Question 39.
The volume of HCl gas weighing 73 g at STP is ……….
(a) 2.24 x 10-2 m3
(b) 4.48 x 10-2 m3
(c) 4.48 x 102m3
(d) 2.24 x 102m3
Answer:
(b) 4.48 x 10-2 m3
Hint:
HCl = Molar mass = 1 + 35.5 = 36.5 g
36.5 g of HCl occupies 2.24 x 10-2m3
∴ 73 g of HCl at STP will occupy \(\frac{2.24 \times 10^{-2}}{36.5}\) x 73 = 4.48 x 10-2m3

Question 40.
The molar volume of 22 g of CO2 is ……….
(a)2.24 x 10-2m3
(b)4.48 x 10-2m3
(c) 1.12 x 10-2m3
(d)2.24 x 10-2m3
Answer:
(c) 1.12 x 10-2m3
Hint:
CO2 = Molar’mass of 12 + 32 = 44 g
44 g of CO2 occupies molar volume = 2.24 x 10-2m3
∴ 2g of CO2 will occupy = \(\frac{2.24 \times 10^{-2}}{44}\) x 2= 1.12 x 10-2 L

Question 41.
The equivalent mass of Aluminium is ……….
(a) 27
(b) 13.5
(c) 54
(d) 9
Answer:
(d) 9

Question 42.
The equivalent mass of HSO4 is ……….
(a) 98
(b) 97
(c) 48
(d) 96
Answer:
(a) 98
Hint:
HSO4 = Molar mass= 1 + 32 + 64 + 1 = 98
Equivalent mass = Molar mass / 1 = 98

Question 43.
The equivalent mass of NaCl is ……….
(a) 40
(b) 58.5
(c) 35.5
(d) 23
Answer:
(b) 58.5
Hint:
NaCl = Salt Molar mass 23 + 35.5 = 58.5
Equivalent mass of Salt = Molar mass of Salt.

Question 44.
Match the List-I and List-lI using the correct code given below the list.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 45.
Match the List-I and List-Il using the correct code given below the list.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 46.
Consider the following statements ……….
(i) Empirical formula shows the actual number of atoms of different elements in one molecule of the compound.
(ii) Ozone is a diatomic molecule.
(iii) Gases are easily compressible.
Which of the above statement is/are not correct?
(a) (i), (ii), (iii)
(b) (i) & (ii)
(c) (ii) & (iii)
(d) (iii) only
Answer:
(b) (i) & (ii)

Question 47.
How many molecules of hydrogen is required to produce 4 moles of ammonia?
(a) 15 moles
(b) 20 moles
(c) 6 moles
(d) 4 moles
Answer:
(c) 6 moles
Hint:
3H2 + N2 → 2NH3
To get 2 moles of ammonia, 3 mole of H2 is required.
To get 4 moles of ammonia = \(\frac {3}{2}\), x 4
= 6 moles of H2 is required.

Question 48.
The number of moles of oxygen required to prepare 1 mole of water is …………..
(a) I mole
(b) 0.5 mole
(c) 2 moles
(d) 0.4 mole
Answer:
(b) 0.5 mole
Hint:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations - 147
0.5 mole of oxygen is required to prepare 1 mole of H2O.

Question 49.
How much volume of CO2 is produced when 50 g of CaCO3 is heated strongly?
(a) 2.24 x 10-2 m3
(b) 22.4
(c) 11.2 L
(d) 22400 cm3
Answer:
(c) 11.2 L
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations - 148
100 g of CaCO3 produces 22.4 litres of CO2.
50 g of CaCO3 will produce = \(\frac{22.4}{100} \times 50\) = 11.2 litres

Question 50.
Which one of the following is not a redox reaction?
(a) Rusting of iron
(b) Extraction of metal Na
(c) Electroplating
(d) Aluminothermic process
Answer:
(a) Rusting of Iron

Question 51.
In the reaction 2 AuCl3 + 3 SnCl2 → 2 Au + 3 SnCl4 which is an oxidising agent?
(a) AuCl3
(b) Au
(c) SnCl2
(d) Both AuCl3 and SnCl2
Answer:
(a) AuCl3
Hint:
AuCl3 undergoes reduction. So, it is an oxidising agent.

Question 52.
Identify the compound formed during the rusting of iron.
(a) Fe2O3
(b) Fe2O3. x H2O
(c) FeO. x H2O
(d) FeO
Answer:
(b) Fe2O3. x H2O – Hydrated iron oxide is rust.

Question 53.
The oxidation state of a substance in its elementary state is equal to ………..
(a) -1
(b) -2
(c) zero
(d) charge of the ion
Answer:
(c) zero

Question 54.
The oxidation number of fluorine in all its compounds is equal to ………….
(a) -1
(b) +1
(c) – 2
(d) +2
Answer:
(a) -1

Question 55.
Consider the following statements.
(i) The sum of the oxidation number of all the atoms in neutral molecule is equal to zero.
(ii) Fluorine has an oxidation number +1 in all its compounds. .
(iii) The oxidation number of a substance in its elementary state is equal to zero.
Which of the above statement is/are not correct?
(a) (i), (ii) & (iii)
(b) (ii) & (iii)
(c) (i) only
(d) (ii) only
Answer:
(d) (ii) only

Question 56.
The oxidation number of Cr in K2Cr2O7 is ………….
(a) +4
(b) + 6
(c)  O
(d) + 7
Answer:
(b) + 6
K2Cr2O2
2 + 2x – 140
2x – 12 = 0
2x = + 12
x = + 6

Question 57.
The oxidation number of N in NH4 ion is ……….
(a) +4
(b) + 3
(c) – 3
(d) – 4
Answer:
(c) – 3
NH2+
x + 4 = + 1
x = + 1 – 4
x = – 3

Question 58.
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s). In this reaction, which gets oxidised?
(a) Cu2+
(b) Zn2+
(c) Zn
(d) Zn, Cu2+
Answer:
(c) Zn
Hint:
Zn → Zn2+ + 2e (loss of electron = oxidation)
Zn gets oxidised

Question 59.
Which one of the following is an example for disproportionation reaction?
(a) CuSO4 + Zn → ZnSO4 + Cu
(b) 2KClO3 → 2KCI + 3O2
(c) PCl5 → PCl3 + Cl2
(d) 4H3PO3 → 3H3PO4 + PH3
Answer:
(d) 4H3PO3 → 3H3PO4 + PH3 (Auto oxidation and reduction reaction)

Question 60.
The number of molecules in 40 g of sodium hydroxide is ……….
(a) 6.023 x 1023
(b) 3.0115 x 1023
(c) 6.023 x 1023
(d) 2 x 6.023 x 1023
Answer:
(c) 6.023 x 1023
Sodium hydroxide = NaOH = 23 + 16+ 1 =40 g
40g = 1 mole = 6.023 1023

Question 61.
The mass of one molecule of AgCl in grams is ……….
(a) 108 g
(b) 143.5 g
(c) 35.5 g
(d) 243.5 g
Answer:
(b) 143.5 g
Hint:
Mass of AgCl = 108 + 35.5 = 143.5 g.

Question 62.
The empirical formula of Alkene is ……….
(a) CH
(b) CH2
(c) CH3
(d) CH3O
Answer:
(b) CH2
Hint:
Alkene CnH2n Molecular formula
E.F. = M.F./2
∴ Empirical formula = CH2

Question 63.
22 g of a gas occupies 11.2 litres of volume at STP. The gas is ……….
(a) CH4
(b) NO
(c) CO
(d) CO2
Answer:
(d) CO2
Hint:
22 g of a gas occupies 11.2 litres.
11.2 liters is occupied by 22 g of a gas.
∴ Molar volume 22.4 liter will be occupied by \(\frac {22}{11.2}\) x 22.4 = 44 g
∴ The gas is CO2.

Question 64.
The number of moles of H2 in 2.24 liter of hydrogen gas at STP is ……….
(a) 1
(b) 0.1
(c) 0.01
(d) 0.001
Answer:
(b) 0.1
Hint:
22.4 liter = 1 molar volume = 1 mole.
∴ 2.24 liter = \(\frac {1}{22.4}\) x 2.24 = 01 mole

Question 65.
How many molecules are present in 32 g of methane?
(a) 2 x 6.023 x 1023
(b) 6.023 x 1023
(c) 6.023 x 1023
(d) 3.011 x 1023
Answer:
(a) 2 x 6.023 x 1023
Hint:
Methane (CH4) – Molar mass = 12 + 4 = 16g.
16 g contains 6.023 x 1023 molecules.
∴ 32 g of methane will contain = \(\frac{6.023 \times 10^{23}}{16} \times 32^{2}\) = 2 x 6.023 x 1023

Question 66.
The empirical formula of glucose is ……….
(a) CH
(b) CH2O
(c) CH2O2
(d) CHO
Answer:
(b) CH2O
Hint:
Glucose = Molecular formula = C6H12O6
Empirical formula = \(\frac { Molecular formula}{6}\) = \(\\frac{\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}{6}\) = CH2O

Question 67.
How many moles of water is present in I L of water?
(a) 1
(b) 18
(c) 55.55
(d) 5.555
Answer:
(c) 55.55
Hint:
1 Liter of water = 1000 g.
Molar mass of water 18 g
Number of moles = \(\frac {Mass}{Molar mass}\) = \(\frac {1000}{18}\) = 55.55 moles

Question 68.
How many moles of Hydrogen atoms are present in 1 mole of C2 H6 ?
(a) 18 moles
(b) 6 moles
(c) 3 moles
(d) 1 mole
Answer:
(b) 6 moles
Hint:
C2H6 contains 6H atoms. 6 moles.

Question 69.
The molar mass of Na2SO4 is ……….
(a).129
(b) 142
(c) 110
(d) 70
Answer:
(b) 142
Hint:
Na2 SO4 = Molar mass
= (23 x 2) + (32 x 1) + (16 x 4)
= 46 + 32 + 64 = 142

Question 70.
Match the List-I with List-Il using the correct code given below the list.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 71.
Ore mole of CO2 contains ………….
(a) 6.02 x 1023 atoms of C
(b) 3 g of CO2
(c) 6.02 1023 atoms of O
(d) 18.1 x 1023 molecules of CO2
Answer:
(a) 6.02 x 1023 atoms of C

Question 72.
5.6 liters of oxygen at STP is equivalent to ……….
(a) 1 mole
(b) 1/4 mole
(c) 1/8 mole
(d) 1/2 mole
Answer:
(b) 1/4 mole
Hint:
22.4 litres of O2 = 1 mole
∴ 5.6 litres of O2 = \(\frac {1}{22.4}\) x 5.6 = 0.25 mole = 1/4 mole.

Question 73.
How many grams are contained in 1 gram atom of Na?
(a) 13 g
(b) 1 g
(c) 23 g
(d) 1/23 g
Answer:
(c) 23 g
Hint:
1 gram atom of Na
Na = Atomic mass 23 g (or) 23 amu
1 gram atom of Na = 1 mole = 23 g.

Question 74.
12 g of Mg will react completely with an acid to give ……….
(a) 1 mole of O2
(b) 1/2 mole of H2
(c) I mole of H2
(d) 2 mole of H2
Answer:
(b) 1/2 mole of H2
Hint:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
∴ 12 g of Mg \(\frac {1}{2}\) mole of H2O

Question 75.
which of the following has the highest mass?
(a) I g atom of C
(b) 1/2 mole of CH4
(c) 10 ml of water
(d) 3.011 x 1023 atoms of oxygen
Answer:
(a) I g atom of C
(a) 1 g atom of C = 12 g
(b) 1/2 mole of CH4 = \(\frac {12 + 4}{2}\) = 8 g.
(c) 10 ml of water (H2O) = 1 x 10 = 10 g
(d) 3.011 x 1023 atoms of oxygen = 0.5 mole of oxygen = 8 g

Question 76.
The empirical formula of sucrose is ……….
(a) CH2O
(b) CHO
(c) C12H22O11
(d) C(H2O)2
Answer:
(a) CH2O
Hint:
Sucrose Molecular formula = C12H22O11
E.F.= \(\frac{\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}}{12}\) = CH2O

Question 77.
The number of grams of oxygen in 0.10 mol of Na2CO3. 10H2O is ………..
(a) 20.8 g
(b) 18 g
(c) 108 g
(d) 13 g
Answer:
(a) 20.8 g
Hint:
Na2CO2.10H2O = 1 mole
1 mole of Na2CO3. 10H2O contains 13 oxygen atoms.
Mass of 13 oxygen atoms = 13 x 16 = 208
1 mole of Na2CO3.10H2O contains 208 g of oxygen.
∴ 0.10 mole of Na2CO3.10H2O contains \(\frac {208}{1}\) x 0.12 = 08 g.

Question 78.
The mass of an atom of nitrogen is ……….
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 79.
Which of the following halogens do not exhibit positive oxidation number in its compounds?
(a) Fluorine
(b) Chlorine
(c) Iodine
(d) Bromine
Answer:
(a) Fluorine

Question 80.
Which of the following is the most powerful oxidising agent?
(a) KMnO4
(b) K2Cr2O7
(c) O3
(d) H2O2
Answer:
(a) KMnO4

Question 81.
On the reaction 2Ag + H2SO4 → Ag2SO4 + 2H2O + SO2. Sulphuric acid acts as ……………
(a) oxidising agent
(b) reducing agent
(c) a catalyst
(d) an acid as well as an oxidant
Answer:
(d) an acid as well as an oxidant

Question 82.
The oxidation number of carboxylic carbon atom in CH3COOH is ……….
(a) + 2
(b) + 4
(c) + 1
(d) + 3
Answer:
(d) + 3
CH3COOH
-3 + 3 + x – 4 + 1
x – 3 = 0
x = + 3
Carboxylic carbon oxidation number = + 3

Question 83.
When methane is burnt in oxygen to produce CO2 and H2O, the oxidation number of carbon changes by ……….
(a) – 8
(b) + 4
(c) zero
(d) + 8
Answer:
(b) + 4
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 84.
The oxidation number of carbon is zero in ……….
(a) HCHO
(b) C12H22O11
(c) C6H12O6
(d) all the above
Answer:
(d) all the above

Question 85.
The oxidation number of Fe in Fe2(SO4)3 is ……….
(a) + 2
(b) + 3
(c) + 2, + 3
(d) O
Answer:
(b) + 3
Fe2(SO4)3
2x – 6 = 0
x = + 3

Question 86.
Among the following molecules in which Chlorine shows maximum oxidation state?
(a) Cl21
(b) KCl
(c) KClO3
(d) Cl2O7
Answer:
(d) Cl2O7
Cl2O7
2x – 14 = 0
2x = + 14
x = + 7

Question 87.
The oxidation number of carbon in CH3 → CH2OH is ………….
(a) + 2
(b) – 2
(e) O
(d) + 4
Answer:
(b) – 2
C2H5OH
2x +6 – 2 = O
2x + 4 = 0
2x = -4
x = -2

Samacheer Kalvi 11th Chemistry Basic Concepts of Chemistry and Chemical Calculations 2 – Mark Questions

I. Write brief answer to the following questions:

Question 1.
State Avogadro’s Hypothesis.
Answer:
It states that ‘Equal volume of all gases under the same conditions of temperature and pressure contain the same number of molecules’.

Question 2.
What is molar volume?
Answer:
Molar volume is the volume occupied by one mole of a substance in the gaseous state at STP. It is equal to 2.24 x 10-2m3 (22.4 L).

Question 3.
The approximate production of Na2CO3 per month is 424 x 106 g while that of methyl alcohol is 320 x 106 g. Which is produced more in terms of moles?
Answer:
Na2CO3 mass = 424 x 106g
Molecular mass of Na2CO3 = (23 x 2) + 12 + (16 x 3)
= 46+ 12 +48
= 106 g
No. of moles of Na2CO3 Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
= 4 x 106 moles
Methyl alcohol mass = 320 x 106 g
Molecular mass of CH3OH = 12 + (1 x 4)+ 16 = 32 g
= 12 + 4 + 16 = 32 g
No. of moles of Methyl alcohol Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
= 10 x 106 moles.
∴ Methyl alcohol is more produced in terms of moles.

Question 4.
Calculate number of moles of carbon atoms ¡n three moles of ethane.
Answer:
Ethane – Molecular formula = C2H6
1 mole of ethane contains 2 atoms of carbon (6.023 x 1023 C)
∴ 3 moles of ethane contains 6 atoms of Carbon.
∴ No. of moles of Carbon atoms = 3 x 6.023 x 1023 Carbon atoms.
= 18.069 x 1023 Carbon atoms.

Question 5.
Find the molecular mass of FeSO4.7H2O
Answer:
Sum of Atomic mass of all elements = Molecular mass
Atomic mass of Fe = 56.0
Atomic mass of S = 32.0
Atomic mass of 4[O] = 64.0
Atomic mass of 14[H] = 14.0
Atomic mass of 7[O] = 112.0 = 278.0
Molecular mass of FeSO4.7H2O = 278 g.

Question 6.
Mass of one atom of an element ¡s 6.66 x 1023 g. How many moles of element are there in 0.320 kg?
Answer:
Mass of one atom of an element = 6.66 x 1023g
No. of moles = \(\frac {Mass}{Molecular mass}\) 3
Molecular mass = Mass of 1 atom x Avogadro number
6.66 x 1023 x 6.023 x 1023
= 6.66 x 6.023 = 40.11318
Number of moles = \(\frac {Mass}{Molecular mass}\) = \(\frac{0.320 \mathrm{kg} \times 10^{3}}{40}\) = 8 moles.

Question 7.
How many moles of glucose are present in 720 g of glucose?
Answer:
Glucose = C6H4O4
Molecular mass of Glucose = (12 x 6) + (1 x 12) + (16 x 6)
= 72 + 12 + 96 = 180
\(\frac {720}{180}\) = 4 moles.

Question 8.
Calculate the weight of 0.2 mole of sodium carbonate.
Answer:
Sodium carbonate = Na2CO3
Molecular mass of Na2CO3 = (23 x 2)+(12 x 1)+(16 x 3)
= 46 + 12 + 48 = 106 g
Mass of 1 mole of Na2CO3 = \(\frac{106 \times 0.2}{1}\) = 21.2 g

Question 9.
What do you understand by the terms acidity and basicity?
Answer:
Acidity:
The number of hydroxyl ions present in one mole of a base is known as the acidity of the base.

Basicity:
The number of replaceable hydrogen atoms present in a molecule of the acid is referred to as its basicity.

Question 10.
Calculate the equivalent mass of bicarbonate ion.
Answer:
Bicarbonate ion = HCO3
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Formula mass of HCO3 = 1 + 12 + 48 = 61
Equivalent mass of HCO3 = \(\frac {61}{1}\) = 61

Question 11.
Calculate the equivalent mass of barium hydroxide.
Barium hydroxide = Ba(OH)2
Molecular mass of Ba(OH)2 = 137 + (16 x 2) + (1 x 2)
= 171.0 g / mol.
Acidity = 2
Equivalent mass of Ba(OH)2 =
= \(\frac {17 1.0}{2}\) = 85.5

Question 12.
Calculate the equivalent mass of hydrated sodium carbonate.
Answer:
Hydrated sodium carbonate = Na2CO3. 10H2O
Molecular mass of Na2CO3. 10H2O = (23 x 2) + (2 x 1) + (16 x 13) + (1 x 20)
= 46 + 12 + 208 + 20 = 286
Equivalent mass of Na2CO3. 10H2O = \(\frac {Molecular mass}{Acidity}\)
= \(\frac {286}{2}\) = 143

Question 13.
What do you understand by the terms empirical formula and molecular formula?
Answer:
Empirical Formula:

  • It is the simplest formula.
  • It shows the ratio of number of atoms of different elements in one molecule of the compound.

Molecular Formula:

  • It is the actual formula.
  • It shows the actual number of different types of atoms present in one molecule of the compound.

Question 14.
Boric acid, H3BO3 is a mild antiseptic and is often used as an eye wash. A sample contains 0.543 mol H3BO3. What is the mass of boric acid in the sample?
Answer:
Molecular mass of H3BO3 = (1 x 3) + (11 x 1) + (16 x 3) = 62
Boric acid sample contains 0.543 mole.
Mass of 0.543 mole of Boric acid = Molecular mass x mole
= 62 x 0.543
= 33.66 g

Question 15.
A compound contains 50% of X (atomic mass 10) and 50% Y (atomic mass 20). Give its molecular formula
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
∴ The Empirical Formula is X3Y
Empirical Formula mass = 20 + 20 = 40
Molecular mass = Sum of atomic mass = 40
n = 1, Molecular formula = (Empirical Formula )n = (X2Y)1 = X2Y.

Question 16.
Calculate the mass of sodium (in kg) present in 95 kg of a crude sample of sodium nitrate whose percentage purity is 70%.
Answer:
Sodium Nitrate = NaNO3
Molecular mass of Sodium Nitrate = 23 + 14 + 48 = 85
100% pure 85 g of NaNO3 contains 23 g of Sodium.
100% pure 95 x 103 g of NaNO3 will contains \(\frac {23}{85}\) x 95 x 103
= 25.70 x 103 g of Sodium.
100% pure NaNO3  contains 25.70 x 103 g of Sodium.
∴ 70% pure NaNO3  will contains = img
= 17990 g (or) 17.99 Kg of Na.

Question 17.
Define matter. What are the types of matter?
Answer:

  • A matter is anything which has mass and occupies space.
  • Matters exist in all three states such as solid, liquid and gas.

Question 18.
Prove that states of matter are inter convertible.
Answer:
States of matter are inter convertible by changing temperature and pressure.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 19.
What is meant by Plasma state? Give an example.
Answer:
Gaseous state of matter at very high temperature containing gaseous ions and free electron is referred to as the Plasma state. e.g. Lightning.

Question 20.
Differentiate an element and an atom.
Answer:

  • An atom is the ultimate smallest electrically neutral, being made up of fundamental particles such as proton, neutron and electron.
  • An element consists of only one type of atoms. Elements are further divided into metals, non-metals, and metaloids.

Question 21.
Distinguish between a molecule and a compound
Answer:
Molecule:

  • A molecule is the smallest particle made up of one or more than one atom in a definite ratio having stable and independent existence.
  • e.g. Na – Mono atomic molecule O2 – Diatomic molecule P4 – Poly atomic molecule

Compound:

  • A molecule which contains two or more atoms of different elements are called a compound molecule.
  • e.g. CO2 – Carbon dioxide CH4 – Methane H2O – Water

Question 22.
Chlorine has fractional average atomic mass. Justify this statement.
Answer:
Chlorine molecule has two isotopes as in 17Cl35 , 17 Cl37 in the ratio of 77 : 23, so when we are calculating the average atomic mass, it becomes fractional.
The average relative atomic mass of Chlorine = \(\frac {(35 x 77) + (37 x 23)}{100}\) = 35.46 amu

Question 23.
Define molecular mass of a substance.
Answer:
Molecular mass of a substance (element or compound) represents the number of times the molecule of that substance is heavier than 1 / 12th of the mass of an atom of C-12 isotope. Molecular mass = 2 x Vapour density

Question 24.
Calculate the molecular mass of Sulphuric acid (H2SO4). Element
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 25.
Define Avogadro Number.
Answer:
Avogadro number is the number of atoms present in one mole of an element or number of molecules present in one mole of a compound. The value of Avogadro number (N) = 6.023 x 1023

Question 26.
Calculate the number of moles present in 60 g of ethane.
Answer:
No. of moles =Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations  = \(\frac {W}{M}\)
Molar Mass of ethane (C2H6) = 24 + 6 = 30
Number of moles in 60 g of ethane = \(\frac {60}{30}\) = 2 moles.

Question 27.
Calculate the equivalent mass of Copper. (Atomic mass of copper = 63.5)
Answer:
Equivalent mass = \(\frac {Atomic mass}{Valency}\)
Equivalent mass of Copper = \(\frac {63.5}{2}\) = 31.75 g eq-1.

Question 28.
Calculate the equivalent mass of (i) Sulphate ion (ii) Phosphate ion.
Answer:
(i) Sulphate ion (SO42-).
Equivalent mass of Sulphate = Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations = \(\frac {32 + 64}{2}\) = \(\frac {96}{2}\) = 48 g eq-1

(ii) Phosphate ion (P043-)
Molar mass of Phosphate ion = Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations = \(\frac {31 + 64}{3}\) = \(\frac {95}{2}\) = 31.6 = 31.6g eq-1

Question 29.
Calculate the equivalent mass of sulphuric acid.
Answer:
Sulphuric acid = H2SO4
Molar mass of Sulphuric acid = 2 + 32 + 64 = 96
Basicity of Sulphuric acid = 2
Equivalent mass of acid = Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations = \(\frac {96}{2}\) = 49g eq-1

Question 30.
How many moles of hydrogen is required to produce 20 moles of ammonia?
Answer:
3H2 + N2 → 2NH3
A per stoichiometric equation,
No. of moles of hydrogen required for 2 moles of ammonia 3 moles
No. of moles of hydrogen required for 20 moles of ammonia = \(\frac {3}{2}\) x 20 = 30 moles.

Question 31.
Calculate the amount of water produced by the combustion of 32 g of methane.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
As per stoichiometric equation,
16 g of methane produces 36 g of H2O
∴ 32 g of methane will produce = \(\frac {36}{16}\) x 32 = 72 g of water.

Question 32.
How much volume of Carbon dioxide is produced when 25 g of calcium carbonate is heated completely under standard conditions?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
100 g of CaCO3 produces 22.4 L of CO2.
∴ 25 g of CaCO3 will produce = \(\frac {22.4}{100}\) x 25 = 5.6 L of CO2.

Question 33.
How much volume of chlorine is required to prepare 89.6 L of HCl gas at STP?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
2 x 22.4 L of HCl is produced by 22.4 L of Cl2.
∴ 89.6 L of HCl will be produced by = img = 89.6 L = 44.8 L of chlorine.

Question 34.
What is meant by limiting reagent?
Answer:
A large excess of one reactant is supplied to ensure the more expensive reactant is completely converted to the desired product. The reactant used up first in a reaction is called the limiting reagent.

Question 35.
On the formation of SF6 by the direct combination of S and F2, which is the limiting reagent? Prove it.
Answer:
SF6 is formed by burning Sulphur in an atmosphere of Fluorine. Suppose 3 moles of S is allowed to react with 12 moles of Fluorine.
S(l) +3F2(g) → SF6(g)
As per the stoichiometric reaction, one mole of S reacts with 3 moles of fluorine to complete the reaction. Similarly, 3 moles of S requires only 9 moles of fluorine.
∴ It is understood that the limiting reagent is Sulphur and the excess reagent is Fluorine.

Question 36.
Mention any 4 redox reaction that takes place in our daily life.
Answer:

  1. Burning of cooking gas, wood
  2. Rusting of iron articles
  3. Electroplating
  4. Galvanic and electrolytic cells

Question 37.
Calculate the oxidation number of underlined elements in the following.

  1. KMnO4
  2. Cr2O72-

Answer:
1. KMnO4
1(+1) + x + 4 (-2) = 0
x – 7 = 0
∴ x = + 7
Oxidation state of Mn = +7.

2. Cr2O72-
2x + 7(-2) = -2
2x – 14 = – 2
2x = +l2
∴ x = + 6
Oxidation state of Cr = +6.

Question 38.
If 10 volumes of H2 gas react with 5 volumes of O2 gas, how many volumes of water vapour would be produced?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Thus 2 volumes of H2 reacts with 1 volume of O2 to produce 2 volumes of H2O(g).
10 volumes of H2 would react with 5 volumes of O2 to produce 10 volumes of H2O(g)
Thus 10 volumes of H2O will be produced.

Question 39.
Which one of the following will have largest number of atoms?

  1. 1 g of Au(s)
  2. 1 g of Na(s)
  3. 1 g of Li(s)
  4. 1 g of Cl2 (g)

Answer:
1. Molar mass of Au = 197 g mol-1.
No: of atoms in 1 g of Au = \(\frac {1}{197}\) x 6.023 x 1023

2. Molar mass of Na = 23 g mol-1.
No of atoms in 1 g of Na = \(\frac {1}{23}\) x 6.023 x 1023

3. Molar mass of Li = 7 g mol-1
No. of atoms in 1 g of Li = \(\frac {1}{7}\) x 6.023 x 1023

4.  Molar mass of Cl2 = 35.5 g mol-1
No. of atoms in 1 g of Cl2 = \(\frac {1}{35.5}\) x 6.023 x 1023

Comparing the number of atoms, the largest number of atoms will be present in 1 g of Li. Since the mass is same in each case, the element with the lowest molar mass would have the largest number of atoms.
∴ Li with lowest molar mass would have the largest number of atoms.

Question 40.
What will be the mass of one 12C atom in g?
Answer:
Molar mass of 12C = 12.00 g mol-1.
∴ Mass of 6.023 x 1023 carbon atom = 12.0 g
∴ Mass of 1 carbon atom = \(\frac{12}{6.023 \times 10^{23}}\) = 1.992 x 10 g.

Question 41.
Justify the following reaction is a redox reaction.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
In the above reaction, oxygen is removed from CuO. So CuO gets reduced. Oxygen is added to H2 to form water. So H2 gets oxidised. i.e. In CuO, oxidation number of Cu +2 is reduced to O whereas in H2, oxidation of H2 O is increased to + 1. So the above reaction is a redox reaction.

Samacheer Kalvi 11th Chemistry Basic Concepts of Chemistry and Chemical Calculations 3 – Mark Questions

Question 1.
Distinguish among the different physical states of matter.
Answer:
Differences among three physical states of matter (solid, liquid and gas) are as follows

S.No. Properties Solid Liquid Gas
1. Volume Definite Definite No definite
2. Shape Definite No definite No definite
3. Molecular arrangement Very closely packed Loosely packed Very loosely packed
4. Freedom of movement Not much freedom Move around and better than solid Move easy and fast
5. Compressibility Non compressible Less compressible Easily compressible

Question 2.
Define equivalent mass of a salt.
Answer:
Equivalent mass of a salt:
It is defined as the number of parts by mass of the salt that is produced by the neutralization of one equivalent of an acid by a base. Therefore the equivalent mass of the salt is equal to its molar mass.

Question 3.
How much copper can be obtained from 100 g of anhydrous copper sulphate?
Answer:
Anhydrous copper sulphate = CuSO4
Molecular mass of CuSO4 = 63.5 + 32 + (16 x 4)
= 63.5 + 32 + 64
= 159.5 g
159.5 g of CuSO4 contains 63.5 g of copper.
∴ 100 g of CuSO4 contains \(\frac{6.35}{159.5}\) x 100 = 0.39811 x 100 = 39.81 g of Copper.

Question 4.
Calculate the equivalent mass of hydrated ferrous sulphate.
Answer:
Hydrated ferrous sulphate = FeSO4.7H2O
Ferrous sulphate – Reducing agent
Ferrous sulphate reacts with an oxidising agent in acid medium according to the equation.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
16 parts by mass of oxygen oxidised 304 g of FeSO4.
8 parts by mass of oxygen will oxidise \(\frac{304}{16}\) x 8 parts by mass of FeSO4.
= 152
Equivalent mass of Ferrous sulphate (Anhydrous) = 152
Equivalent mass of crystalline Ferrous sulphate FeSO4 .7H2O = 152 + 126 = 278

Question 5.
Give difference between empirical and molecular formula
Answer:
Empirical Formula:

  • Empirical formula is the simplest formula.
  • It shows the ratio of number of atoms of different elements in one molecule of the compound.
  • It is calculated from the percentage of composition of the various elements in one molecule.
  • For example, Empirical formula of Benzene = CH.

Molecular Formula:

  • Molecular Formula is the actual formula.
  • It shows the actual number of different types of atoms present in one molecule of the compound.
  • It is calculated from the Empirical formula Molecular Formula = (Empirical formula)n
  • Molecular formula of Benzene = C6H6.

Question 6.
A sample of hydrated copper sulphate is heated to drive off the water of crystallization, cooled and reweighed 0.869 g of CuSO4 aH2O gave a residue of 0.556 g. Find the molecular formula of hydrated copper sulphate.
Answer:
0.869 g of CuSO4.aH2O gave a residue of 0.556 g of Anhydrous CuSO4.
∴ Weight of a H2O molecule = 0.869 – 0.556 = 0.313 g
Molecular weight of H2O = (1 x 2) + 16 = 2 + 16 = 18
No. of moles of water = \(\frac{Mass}{Molecular mass}\)
CuSO4.5H2O – Molecular mass = 63.5 + 32 + 64 + 90 = 249.5 g
249.5 g of CuSO4.5H2O on heating gives 159.5 g of CuSO4.
0.869 g of CuSO4.aH2O on heating gives = \(\frac{159.5}{249.5}\) x 0.869
= 0.556 g of anhydrous CuSO4 ∴ a = 5
The molecular formula of hydrated copper sulphate = CuSO4.5H2O

Question 7.
Balance by oxidation number method: Mg + HNO3 → Mg(NO3)2 + NO2 + H2O
Step – 1
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Step – 2
Mg + 2HNO3 -» Mg(NO3)2 + NO2 + H2O
Step – 3
To balance the number of oxygen atoms and hydrogen atoms 2HNO3 is multiplied by 2.
Mg + 4HNO3 → Mg(NO3)2 + 2NO2 + H2O
Step – 4.
To balance the number of hydrogen atoms, the H2O molecule is multiplied by 2.
Mg + 4HNO3 → Mg(NO3)2 + 2NO2 + 2H2O

Question 8.
Explain about the classification of matter.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 9.
Calculate the mass of the following atoms in amu,
(a) Helium (mass of He = 6.641 x 10-24 g)
(b) Silver (mass of Ag = 1.790 x 10-22 g)
1 amu = 1.66056 x 10-24
Answer:
(a) The mass of Helium atom in amu = \(\frac{6.641 \times 10^{-24}}{1.66056 \times 10^{-24}}\) = 3.9992 amu.
(b) The mass of Silver atom in amu = \(\frac{1.790 \times 10^{-22}}{1.66056 \times 10^{-24}}\) = 107.79 amu.

Question 10.
Calculate the number of atoms present in 1 Kg of gold.
Answer:
The atomic mass of Gold = 197 g mol-1.
197 g of gold contains 6.023 x 1023 atoms of gold.
∴ 1000 g of gold will contain = \(\frac{1000 \times 6.023 \times 10^{23}}{197}\)
= 3.055 x 1024 atoms of Gold.

Question 11.
Calculate the molar volume of 146 g of HCl gas and the number of molecules present in it.
Answer:
Molar mass of HCl = 36.5 g
The molar volume of 36.5 g (1 mole) of HCl = 2.24 x 102 m3.
∴ The volume of 146 g (4 moles) of HCl = \(\frac{2.24 \times 10^{-2}}{36.5}\) x 146
= 8.96 x 10 m3
No. of molecules in 146 g of HCl = 4 N
= 4 x Avogadro Number
= 4 x 6.023 x 1023
= 24.092 x 1023
= 2.4092 x 1024 molecules.

Question 12.
Calculate the molar mass of 20 L of gas weighing 23.2 g at STP.
Molar mass =Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Molar volume at STP = 2.24 x 10-2 m3 = 22.4 L (or) 22400 cc.
Molar mass of the gas at STP = \(\frac{23.2 x 22.4}{20}\) = 25.984 g.

Question 13.
0.6 g of a metal gives on oxidation 1 g of its oxide. Calculate its equivalent mass.
Answer:
Mass of metal = 0.6
Mass of metal oxide = 1 g
Mass of oxygen = 1 – 0.6 = 0.4 g
0.4 g of oxygen combines with 0.6 g of metal.
∴ 8 g of oxygen will combine with = \(\frac{0.6}{0.4}\) x 8
Equivalent mass of the metal = 12 g eq-1.

Question 14.
How would you calculate the equivalent mass of anhydrous oxalic acid and hydrated oxalic acid.
Answer:
In acid medium,
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
16 g of oxygen is used for oxidation of 90 g of oxalic acid.
∴ 8 g of oxygen will oxidize = \(\frac{90}{16}\) x 8 = 45 g eq-1.
Equivalent mass of Anhydrous oxalic acid = 45 g eq-1
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
= \(\frac{126}{2}\) = 63 g eq-1.

Question 15.
A compound on decomposition in the laboratory produces 24.5 g of nitrogen and 70 g of oxygen. Calculate the empirical formula of the compound.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
the empirical formula is N2O5

Question 16.
What is the steps involve in the calculation of molecular formula from empirical formula?
Answer:
Molecular mass and empirical formula are used to deduce molecular formula of the compound.
Steps to calculate molecular formula:

  • if Empirical formula is found out from the percentage composition of elements
  • Empirical formula mass can be found from the empirical formula
  • Molecular mass is found out from the given data
  • Molecular formula = (Empirical formula)n
  • where, n = Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 17.
What is combination reaction? Give example.
When two or more substances combine to form a single substance, the reactions are called combination reactions.
A + B → C
Example:
2 Mg + O2 → 2MgO

Question 18.
What is decomposition reaction? Give two examples.
Answer:
Chemical reactions in which a compound splits up into two or more simpler substances are called decomposition reaction.
AB → A + B
Example:
2KCO3 → 2KCl + O2
PCl5 → PCl3 + Cl2

Question 19.
What is displacement reactions? Give its types. Explain with example.
Answer:
The reactions in which one ion or atom in a compound is replaced (or substituted) by an ion or atom of the other element are called displacement reactions.
AB + C → AC + B

Example:
Metal displacement
CuSO4 + Zn → ZnSO4 + Cu

Example:
Non-metal displacement
2KBr + Cl2 → 2KCl + Br2

Question 20.
What is disproportionation reactions? Give example.
Answer:
The reactions in which an element undergoes simultaneously both oxidation and reduction are called as disproportionation reactions.
Example:
P4 + 3NaOH + 3H2O → PH3 + 3NaH2PO2
2HCHO + H2O → CH3OH + HCOOH

Question 21.
What are competitive electron transfer reaction? Give example.
Answer:
These are the reactions in which redox reactions take place in different vessels and it is an indirect redox reaction. There is a competition for the release of electrons among different metals.
Example:
Zn releases electrons to Cu and Cu releases electrons to Silver and so on.
Zn(s) + Cu2+ → Zn2+(aq) + Cu(s) (Here Zn – oxidised; Cu2+ – reduced)
Cu(s) + 2Ag+ → Cu2+(aq) + 2Ag(s) (Here Cu – oxidised; Ag+ – reduced)

Question 22.
Balance the following equation using oxidation number method.
Answer:
1. S + HNO3 → H2SO4 + NO2 + H2O
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
2. S + 6HNO3 → H2SO4 + NO2 + H2O
3. Balance the equation (except O and H)
S + 6HNO3 → H2SO4 + 6NO2 + H2O
4. Balance O atoms by adding 2H2O
5 + 6HNO3 → H2SO4 + 6NO2 + 2H2O

Question 23.
Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% oxygen by mass.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
the empirical formula is Fe2O3

Question 24.
In three moles of ethane (C2H6) calculate the following:

  1. Number of moles of carbon atoms.
  2. Number of moles of hydrogen atoms.
  3. Number of molecules of ethane.

Answer:

(1) 1 mole of C2H6 contains 2 moles of Carbon atoms.
∴ 3 moles of C2H6 will have 6 moles of Carbon atoms.
(2) 1 mole of C2H6 contains 6 moles of Hydrogen atoms.
∴ 3 moles of C2H6 will have 18 moles of Hydrogen atoms.
(3) 1 mole of C2H6 contains 6.023 x 1023 number of molecules.
∴ 3 moles of C2H6 will contain 3 x 6.023 x 1023molecules.

Question 25.
Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction.
4HCl(aq) + MnO2(s) → 2H2O(l) + MnCl2(aq) + Cl2(aq)
How many grams of HCl react with 5.0 g of manganese dioxide? (Atomic mass of Mn = 55 g).
Answer:
1 mole of MnO2 = 55 + 32 = 87 g.
87 g of MnO2 reacts with 4 moles of HCl. i.e. = 4 x 36.5 = 146 g of HCl.
∴ 5 g of MnO2 will react with \(\frac{146}{87}\) x 5.0 = 8.40 g.

Question 26.
The density of water at room temperature is 1.0 g/ml. How many molecules are there in a drop of water if its volume is 0.05 ml?
Answer:
Volume of drop of water = 0.05 ml
Mass of a drop of water = Volume x Density
= 0.05 ml x 1.0 g / ml.
= 0.05 g
Molar mass of water (H2O) = 18 g
18 g of water = 1 mole
0. 05 g of water = \(\frac{1}{18}\) x 0.05 = 0.0028 mol.
No. of molecules present in one mole of water = 6.023 x 1023
No. of molecules present in 0.0028 mole of water = \(\frac{6.023 \times 10^{23} \times 0.0028}{1}\) = 1.68 x 1021 water molecules

Question 27.
Balance the following equation by oxidation number method. MnO4 + Fe2+ → Mn 2+ + Fe3+ (Acidic medium)
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
MnO4 + 5e → Mn2+ ……….(1)
Fe2+ → Fe3+ + e ……….(2)
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
to balance O and H atoms H2O and H+ are added.
MnO4 + 5Fe2+ + 8H+ → 5Fe3+ + Mn2+ + 4H2O

Samacheer Kalvi 11th Chemistry Basic Concepts of Chemistry and Chemical Calculations 5-Mark Questions

I. Answer the following questions in detail
Question 1.
Define the following (a) equivalent mass of an acid (b) equivalent mass of a base (c) equivalent mass of an oxidising agent (cl) equivalent mass of a reducing agent.
Answer:
(a) Equivalent mass of an acid:
Equivalent mass of an acid is the number of parts by mass of the acid which contains 1.008 part by mass of replaceable hydrogen atom.
Equivalent mass of an acid = Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

(b) Equivalent mass of a base:
It is defined as the number of parts by mass of the base which contains one replaceable hydroxyl ion or which completely neutralizes one gram equivalent of an acid.
Equivalent mass of a base = Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

(c) Equivalent mass of an oxidising agent:
It is defined as the number of parts by mass of an oxidising agent which can furnish 8 parts by mass of oxygen for oxidation.

(d) Equivalent mass of a reducing agent:
It is defined as the number of parts by mass of the reducing agent which is completely oxidised by 8 parts by mass of oxygen or with one equivalent of any oxidising agent.

Question 2.
Calculate the percentage composition of the elements present in lead nitrate. How many Kg of 02 can be obtained from 50 kg of 70% pure lead nitrate?
Answer:
Lead nitrate = Pb (NO3)2
Molecular mass of lead nitrate = 207 + (14 x 2) + (16 x 6)
= 207 + 28 + 96 = 331 g / mol.
331 g of lead nitrate contains 96 g of oxygen.
∴ 50 x 103 g of lead nitrate will contain \(\frac {96}{331}\) x 50 x 103
= 14501.5 g
= 14.501 Kg of oxygen.
100 % pure lead nitrate contains 14.501 Kg of oxygen.
70 % pure lead nitrate will contain = \(\frac {14.501}{100}\) x 70 = 10.15 Kg of oxygen.
.’. 70 % pure lead nitrate will contain 10.15 Kg of oxygen.

Question 3.
Determine the empirical formula of a compound containing K = 24.15%, Mn = 34.77% and rest is oxygen.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
empirical formula of a compound = KMnO4

Question 4.
Write the steps to be followed for writing empirical formula.
Answer:
Empirical formula shows the ratio of number of atoms of different elements in one molecule of the compound.
Steps for finding the Empirical formula:
The percentage of the elements in the compound is determined by suitable methods and from the data collected; the empirical formula is determined by the following steps.

  1. Divide the percentage of each element by its atomic mass. This will give the relative number of atoms of various elements present in the compound.
  2. Divide the atom value obtained in the above step by the smallest of them so as to get a simple ratio of atoms of various elements.
  3. Multiply the figures so obtained, by a suitable integer if necessary in order to obtain whole number ratio.
  4. Finally write down the symbols of the various elements side by side and put the above numbers as the subscripts to the lower right hand of each symbol. This will represent the empirical formula of the compound.
  5. Percentage of Oxygen = 100 – Sum of the percentage masses of all the given elements.

Question 5.
An organic compound was found to contain carbon = 40.65%, hydrogen = 8.55% and Nitrogen = 23.7%. Its vapour density was found to be 29.5. What is the molecular formula of the compound?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Empirical formula = C2H5NO
Molecular mass = 2 x vapour density = 2 x 29.5 = 59.0
Empirical formula mass of C4H5NO = 24 + 5 + 14 + 16 = 59
n = Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations  = \(\frac {59}{59}\) = 1
Empirical formula mass 59
.’. Molecular formula = (Empirical formula)n
= (C2H5NO)1
Molecular formula = C2H5NO

Question 6.
Calculate the empirical and molecular formula of a compound containing 32% carbon, 4% hydrogen and rest oxygen. Its vapour density is 75.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Empirical formula = C2H3O3
Empirical formula mass = 24 + 3 + 48 = 75
Molecular mass = 2 x Vapour density = 2 x 75 = 150
n = Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations  = \(\frac {150}{75}\) = 2
Molecular formula = (Empirical formula)n
Molecular formula = C2H3O3 x 2
Molecular formula = C4H6O6

Question 7.
Explain the different types of redox reactions with example.
Answer:
Redox reactions are classified into the following types:
(1) Combination reactions:
When two or more substances combine to form a single substance, the reactions are called combination reactions.
Example:
2Mg + O2 → 2MgO

(2) Decomposition reactions:
Chemical reactions in which a compound splits up into two or more simpler substances are called decomposition reaction.
Example:
2KClO3 → 2KCl + 3O2

(3) Displacement reactions:
The reactions in which one ion or atom in a compound is replaced by an ion or atom of the other element are called displacement reactions.
Example:
CuSO4 + Zn → ZnSO4 + Cu

(4) Disproportionation reactions:
The reactions in which an element undergoes simultaneously both oxidation and reduction are called as disproportionation reactions.
Example:
2HCHO + H2O → CH3OH + HCOOH

(5) Competitive Electron transfer reactions:
These are the reactions in which redox reactions take place in different vessels and it is an indirect redox reaction. There is a competition for the release of electrons among different metals.
Example:
Zn(s) + Cu2+ → Zn2+(aq) + Cu(s)  Here Zn – oxidised; Cu2+ reduced

Question 8.
Write the steps to be followed while balancing redox equation by oxidation number method.
Answer:
Oxidation number method:
This method is based on the fact that
Number of electrons lost by atoms = Number of electrons gained by atoms

Steps to be followed while balancing Redox reactions by Oxidation Number method:

  1. Write skeleton equation representing redox reaction
  2. Write the oxidation number of atoms undergoing oxidation and reduction.
  3. Calculate the increase or decrease in oxidation numbers per atom.
  4. Make increase in oxidation number equal to decrease in oxidation number by multiplying the formula of oxidant and reductant by suitable numbers.
  5. Balance the equation atomically on both sides except O and H atoms.
  6. Balance oxygen atoms by adding required number of water molecules to the side deficient
    in oxygen atoms.
  7. Add required number of H+ ions to the side deficient in hydrogen atom if the reaction is in acidic medium.
  8. For reactions in basic medium, add H2O molecules to the side deficient in hydrogen atoms and simultaneously add equal number of OH ions on the other side of the equation.
  9. Finally balance the equation by cancelling common species present on both sides of the equation.

Question 9.
Balance the following equation by oxidation number method:
Answer:
K2Cr2O7 + KCl + H2SO4 → KHSO4 + CrO2Cl 2 + H2O
Step – 1
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Step – 2
K2Cr2O7 + KCl + H2SO4 → KHSO4 + 2CrO2C2 + H2O

Step 3.
To balance Cl atom, KC1 is multiplied by 4
K2Cr2O7 + 4KCl + H2SO4 → 2CrO2Cl2 + KHSO2 + H2O

Step 4.
To balance K atom, KHSO4 is multiplied by 6.
K2Cr2O7 + 4KCl + H2SO4 → 2CrO2Cl2 + 6KHSO4+ H2O

Step 5.
To balance O and H atoms, HSO4 is multiplied by 6, H20 is multiplied by 3.
Answer:
K2Cr2O7 + 4KCl + 6H2SO4 → 2CrO2Cl2 + 6KHSO4 + 3H2O

(2) P + HNO3 → H3PO4 + NO2 + H2O
Step – 1
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Step 2.
P + 5HNO3 → H3PO4 + 5NO2 + H2O

(3) CuO + NH3 → Cu + N2 + H2O
Step 1.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Step 2.
3CuO + 2NH3 → 3Cu + N2 + H2O

Step 3.
To balance O and N, water is multiplied by 3.
3CuO + 2NH3 → 3Cu + N2 + 3H2O

(4) Zn + HNO3 →Zn(N03)2 + NH4NO3 + H2O
Step – 1.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Step 2.
4Zn + HNO3 → 4Zn(NO3)3 + NH4NO2 + H2O

Step 3.
To balance N, HNO3 is multiplied by 10
4Zn + 10HNO3 → 4Zn(NO3)2 + NH4NO3 + H3O

Step 4.
To balance oxygen, H2O is multiplied by 3
4Zn + 10HNO3 → 4Zn(NO3)2 + NH4NO3 + 3H3O

Question 10.
Balance the following equation by ion-electron method In acidic medium.
Answer:
(i) S2O32- + I2 → S2O42- + SO2 + I
Oxidation half reaction:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Reduction half reaction:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
To balance, SO2 is added on RHS of the equation.
S2O32- + I2 → S2O42- + 2I + SO2
To balance oxygen atom, S2O32- and SO2 is multiplied by 2.
2S2O32- + I2 → S2O42- + 2I – + 2SO2

(2) Sb3+ + MnO4 → Sb5+ + Mn2+
Oxidation half reaction:
Sb3+ → Sb5+ + 2e ……….(1)
Reduction half reaction:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
In equation (2), H2O is added on L.HS to balance oxygen atom.
MnO2 + 5e → Mn2+ + 4H2O ………(3)
To balance Hydrogen atoms, H’ is added on RHS.
MnO4 + 5e + 8H+ → Mn2+ + 4H2O ………(4)
Equation (4) is multiplied by 2 and equation (I) is multiplied by 5 to equalise the electrons gained and electrons lost.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

(3) MnO4 + I → MnO2 + I2
Oxidation half reaction:
2I + 2e → I ………..(1)
Reduction half reaction:
MnO4  → MnO2 + e ………..(2)                                    
Equation (1)is multiplied by 3
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Equation (2)is multiplied by 2.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
To balance oxygen and hydrogen atoms, H+ is added on RHS and H2O is added on LHS.
2MnO4 + 6I + 4H+ → 2MnO2 + 2I2 + 2H2O

In acidic medium
(4) MnO4 + Fe2+ → Mn2+ + Fe3+
Oxidation half reaction:
Fe2+ → Fe3+ + e ……..(1)
Reduction half reaction:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
To balance oxygen, H+ is added on RHS and H2O is added on LHS.
MnO4 + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O

(5) Cr(OH)4 + H2O2 → CrO4
Oxidation half reaction:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Reduction half reaction:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
To balance oxygen and hydrogen atoms, OH and H2O are added.
2Cr(OH)4 + 3H2O2 + 20H → 2CrO2 + 8H2O

Question 11.
(a) Define equivalent mass of an oxidising agent.
(b) How would you calculate the equivalent mass of potassium permanganate?
(a) The equivalent mass of an oxidizing agent is the number of parts by mass which can furnish 8 parts by mass of oxygen for oxidation.
(b) Potassium permanganate is an oxidizing agent.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
80 parts by mass of oxygen are given by 316 g of KMnO4.
.’. 8 parts by mass of oxygen will be furnished by \(\frac {316}{80}\) x 8 = 31.6
Equivalent mass of KMnO4 = 31.6 g eq-1.

Question 12.
(a) Define equivalent mass of an reducing agent.
(b) How would you determine the equivalent mass of Ferrous sulphate?
Answer:
(a) The equivalent mass of a reducing agent is the number of parts by mass of the reducing agent which is completely oxidised by 8 parts by mass of oxygen or one equivalent of any oxidising agent.
(b) Ferrous sulphate is a reducing agent.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
16 parts by mass of oxygen oxidised 304 parts by mass of FeSO4.
∴ 8 parts by mass of oxygen will oxidise \(\frac {316}{80}\) x 8 parts by mass of ferrous sulphate = 152 .
The equivalent mass of ferrous sulphate (anhydrous) = 152.
The equivalent mass of crystalline ferrous sulphate is (FeSO4.7H2O) = 152 + 126 = 278
The equivalent mass of crystalline ferrous sulphate = 278.

Question 13.
A compound on analysis gave the following percentage composition: C = 24.47%, H = 4.07 %, Cl = 7 1.65%. Find out its empirical formula.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
the empirical formula is CH2Cl.

Question 14.
A laboratory analysis of an organic compound gives the following mass percentage composition: C = 60%, H = 4.48% and remaining oxygen.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
∴ the empirical formula is C9H8O4.

Question 15.
An insecticide has the following percentage composition by mass: 47.5% C, 2.54% H, and 50.0% Cl. Determine its empirical formula and molecular formulae. Molar mass of the substance is 354.5 g mol-1
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
The empirical formula is C14H9C15.
Calculation of Molecular formula:
The empirical formula mass (C14H9C15) = (14 X 12) + (9 X 1) + (5 X 35.5)
n = Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations = \(\frac {354.5}{354.5}\) = 1
Empirical formula mass 354.5
Molecular formula = (Empirical formu1a)n
= (C14H9Cl5)1
∴ Molecular formula = C14H9Cl5

Question 16.
An organic fruit smelling compound on analysis has the following composition by mass: C = 54.54%, H = 9.09%, O = 36.36%. Find out the molecular formula of the compound. The vapour density of the compound was found to be 44.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
∴ The empirical formula is C2H4O
Theempirical formula mass(C2H4O) = (12 x 2)+(1 x 4)+(16 x 1) = 44
Molecular mass = 2 x Vapour density = 2 x 44 = 88
n = Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations = \(\frac {88}{44}\) = 2
Molecular formula = (Empirical formula)n
= (C2H4O)2
∴ Molecular formula = C2H4O2

Question 17.
Calculate the percentage composition of the elements present in magnesium carbonate. How many Kg of CO2 can be obtained from 100 Kg of is 90% pure magnesium carbonate.
Molar mass of MgCO3 = 84.32 g mol-1
Percentage of Mg = \(\frac {24}{84.32}\) x 100 = 28.46%
Percentage of C = \(\frac {12}{84.32}\) x 100 = 14.23%
Percentage of O3 = \(\frac {48}{84.32}\) x 100 = 57.0%
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
84.32 g of 100% pure MgCO3 gives 44g of CO2
∴ 100 x 103g of 100% pure MgCO3 gives = \(\frac {44}{84.32}\) x 100 x 103
= 52.182 x 103 g CO2
100% pure MgCO3 gives 52.182 x 103 g CO2
∴ 90% pure MgCO3 will give \(\frac{52.182 \times 10^{3}}{100}\) x 90 = 46963.8 g CO2

Question 18.
Urea is prepared by the reaction between ammonia and carbon dioxide.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
In one process, 637.2 g of NH3 are allowed to react with 1142 g of CO2.
(a) Which of the two reactants is the limiting reagent?
(b) Calculate the mass of(NH4)2CO formed.
(c) how much of the excess reagent in grams is left at the end of the reaction?
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
No. of moles of ammonia = \(\frac{637.2}{17}\) = 37.45 mole
No. of moles of CO2 = \(\frac{1142}{44}\) = 25.95 moles
As per the balanced equation, one mole of CO2 requires 2 moles of ammonia.
∴ No. of moles of NH3 required to react with 25.95 moles of CO2 is = \(\frac {2}{1}\) x 25.95 = 51.90 moles.
∴ 37.45 moles of NH3 is not enough to completely react with CO2 (25.95 moles).
Hence, NH3 must be the limiting reagent, and CO2 is excess reagent.

(b) 2 moles of ammonia produce 1 mole of urea.
∴ Limiting reagent 37.45 moles of NH3 can produce \(\frac {1}{2}\) x 37.45 moles of urea.
= 18.725 moles of urea.
∴ The mass of 18.725 moles of urea = No. of moles x Molar mass
= 18.725 x 60
= 1123.5 g of urea.

(c) 2 moles of ammonia requires 1 mole of CO2.
∴ Limiting reagent 37.45 moles of NH3 will require x 37.45 moles of CO2.
= 18.725 moles of CO2.
∴ No. of moles of the excess reagent (CO2) left = 25.95 – 18.725 = 7.225
The mass of the excess reagent (CO2) left = 7.225 x 44 = 317.9 g CO2.

Question 19.
(a) Define oxidation number.
(b) What are the rules used to assign oxidation number?
Answer:
(a) Oxidation number refers to the number of charges an atom would have in a molecule or an ionic compound, if electrons were transferred completely.
(b) Rules to assign oxidation number:

  • Oxidation number of a substance in its elementary state is equal to zero (H2, Br2, Na)
  • Oxidation number of a mono-atomtic ion is equal to the charge on the Ton (Na4 =+ 1,
  • Oxidation number of hydrogen in a compound is +1 (except hydrides).
  • Oxidation number of hydrogen in metal hydrides is -1 (NaH, CaH2).
  • Oxidation number of oxygen in a compound is -2 (except OF2 and peroxides).
  • Oxidation number of oxygen in peroxides is -1 (He,, Na»,) = I.
  • Oxidation number of oxygen in fluorinated compounds is either +1 or +2.(OF2 = +2, O,F2 + 1).
  • Fluorine has an oxidation number -1 in all ìts compounds.
  • The sum of the oxidation number of all the atoms in neutral molecules is equal to Zero.
  • For all ions, the sum of the oxidation number of all atoms is equal to the charge of the ion.

Question 20.
Balance the following equation by oxidation number method.
C6H6 + O2 → CO2 + H2O
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
(ii) Balance the changes in O.N. by multiplying the oxidant and reductant by suitable numbers
2 C6H6 + 15 O2 → CO2 + H2O

(iii) Balance the equation atomically (except O and H).
2C6H6 + 15 O2 → 12 CO2 + H2O

(iv) Balance O atoms by adding one HzO molecule to the RHS for making the number of molecules of H2O to be 6.
2C6H6 + 15 O2 → 12 CO2 + 6H2O

Question 21.
Balance the following equation by oxidation number method.
KMnO4 + HCl → KCl + MnCl2 + H2O + Cl2
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

(ii) 2KMnO4 + 1OHCl → KCl + MnCl2 + H2O + Cl2

(iii) Balance the equation atomically (except O and H).
2KMnO4 + 1OHCl → 2KCl + 2MnCl2, + H2O + Cl2

(iv) Balance chlorine atoms by adding HC1 and multiplying Cl2 by 5.
2KMnO4 + 16HCl → 2KCl + 2MnCl2 + H2O + 5Cl2

(v) To balance O and H, H2O is multiplied by 8.
2KMnO4  + I6HCl → 2KCl + 2MnCl4 + 8H2O + 5Cl2

Question 22.
Balance the following equation by oxidation number method.
KMnO4 + FeSO4 + H2S04 → K2SO4 + MnSO4 + Fe2(S04)3 + H2O
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
(ii) 2KMnO4 + 10FeSO4 + H2SO4 → K2SO4 + MnSO4 + Fe2(SO4)3+ H2O

(iii) Balance the equation atomically (except O and H) and sulphate ions.
2KMnO4 + 1OFeSO4 + 8H2SO4 → K2SO4 + 2MnSO4 + 5Fe4(SO4)3 + H2O

(iv) Balance O atoms by multiplying H2O by 8.
2KMnO4 + 1OFeSO4 + 8H2SO4 → K2SO4 + 2MnSO4 + 5Fe2(SO4)3 + 8H2O

Question 23.
Balancing of molecular equation in alkaline medium.
MnO2 + O2 + KOH → K2MnO4 + H2O
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

(ii) Balance the changes in O, N, by multiplying the oxidant and reductant by suitable numbers.
4MnO2 + 2O2+ KOH → K2MnO4 + H2O

(iii) Balance the equation atomically (except O and H).
4MnO2 + 2O2 + KOH → K2 MnO4 + H2O

(iv) Balance oxygen and hydrogen atoms by multiplying H20 by 4.
4MnO2 + 2O2 + 8KOH → 4K2MnO4 + 4H2O

Question 24.
Explain the steps involved in ion-electron method for balancing redox reaction.
Answer:
Ion electron method makes use of the Half reactions. Steps involved in this method are,

  1. Write the equation in the net ionic form without attempting to balance it.
  2. Write and locate the oxidation number of atoms undergoing oxidation and reduction from the knowledge of calculation of oxidation number.
  3. Write two half reactions showing oxidation and reduction separately.
  4. Balance oxygen atoms by adding required number of water molecules to the side deficient in oxygen atoms.
  5. Add required number of H+ ions to the side deficient in hydrogen atom if the reaction is in acidic medium
  6. Add electrons to whichever side is necessary to make up the difference in oxidation number.
  7. Add the two half reactions. The resulting equation is a net balanced equation.
  8. For reactions in basic medium, add H2O and hydrogen ion to balance H and O.
  9. Finally balance the equation by cancelling common species present on both sides of the equation.

Question 25.
Write balanced equation for the oxidation of Ferrous ions to Ferric ions by permanganate ions in acid solution. The permanganate ion forms Mn2+ ions under these conditions.
Answer:
Net ionic reaction:
MnO4 + Fe2+ + H+ → Mn2+ + Fe3+
Oxidation half reaction:
Fe2+ → Fe3+ + e ……….(1)
Reduction half reaction:
MnO4 + 5e → Mn2+ ……….(2)
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
To balance O and H, H+ and H2O are added.
MnO4 + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ +4H2O

Question 26.
A flask A contains 0.5 mole of oxygen gas. Another flask B contains 0.4 mole of ozone gas. Which of the two flasks contains greater number of oxygen atoms.
Answer:
Flask A:
1 mole of oxygen gas = 6.023 x 1023 molecules
∴ 0.5 mole of oxygen gas = 6.023 x 1023 x 0.5 molecules
The number of atoms in flask A = 6.023 x 1023 x 0.5 x 2 = 6.023 x 1023 atoms.

Flask B:
1 mole of ozone gas = 6.023 x 1023 molecules
0. 4 mole of ozone gas = 6.023 x 1023 x 0.4 molecules
The number of atoms in flask B = 6.023 x 1023 x 0.4 x 3 = 7.227 x 1023 atoms.
∴ Flask B contains a great number of oxygen atoms as compared to flask A.

Question 27.
(a) Formulate possible compounds of ‘Cl’ in its oxidation state is:
0, – 1, + l, + 3,+ 5, + 7

(b) H2O2 act as an oxidising agent as well as reducing agent where as O3 act as only oxidizing agent. Prove it.
(a) (1) Cl oxidation number O in Cl2.
(2) Cl oxidation number -1 in HCl.
(3) Cl oxidation number +3 in HCl O2.
(4) Cl oxidation number +5 in KClO3.
(5) Cl oxidation number +7 in C12O7.

(b) In H2O2, oxidation number of oxygen is -1 and it can vary from O to -2 (+ 2 is possible in OF2). The oxidation number can decrease or increase, because of this H2O2 can act both oxidising and reducing agent. Ozone (O3 ) only acts as oxidising agent since it decomposes to give nascent oxygen.

Question 28.
The Mn3+ ion is unstable in solution and undergoes disproportionation to give Mn2+, MnO2 and H ion. Write a balanced ionic equation for the reaction.
The skeletal equation is:
Mn3+(aq) → Mn2+(aq) + MnO2(aq) + H+(aq)
Oxidation half equation:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Balance O.N. by adding electrons,
Mn3+(aq) →  MnO2(s) + e
Balance charge by adding 4H ions,
Mn3+(aq) →  MnO2(s) + 4H+(aq) + e
Balance O atoms by adding 2H2O
Mn3+(aq) + 22H2O(l) → MnO2(s) + 4H+(aq) + e ……….(1)
Reduction half equation:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Balance ON. by adding electrons:
Mn3+(aq) + e → Mn2+(aq) …………(2)
Adding Equation (1) and (2), the balanced equation for the disproportionation reaction is
2MH3+(aq) + 2H2O(l) → MnO2(s) + Mn2+(aq) + H+(aq)

Question 29.
Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess chlorine is removed by treating with sulphur dioxide. Present a balanced equation for the reaction for this redox change taking place in water.
The skeletal equation is:
Cl2(aq) + SO2(aq) + H2O(l) → Cl(aq) + SO42-(aq)
Reduction half equation:
Cl2(aq) → Cl2-(aq)
Balance Cl atoms,
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Balance O.N. by adding electrons
Cl2(aq) + 2e → 2Cl + 2e
Oxidation half equation:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Balance O,N. by adding electrons
SO2(aq) → SO42-(aq) + 2e
Balance charge by adding 4H+ ions:
SO2(aq) → SO42-(aq) + 4H+(aq) + 2e
Balance O atoms by adding 2H2O
SO2(aq) + 2H2O(l) → SO42-(aq) + 4H+(aq) + 2e
Adding equation (1) and (2), we have,
Cl2(aq) + SO2(aq) + 2H2O(l) → Cl2(aq) + SO42-(aq) + 4H+(aq)
This represents the balanced redox reaction.

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Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2

11th Maths Exercise 7.2 Question 1.
Without expanding the determinant, prove that11th Maths Exercise 7.2 Samacheer Kalvi Chapter 7 Matrices And Determinants
Solution:
11th Maths Exercise 7.2 Answers Samacheer Kalvi Chapter 7 Matrices And Determinants

11th Maths Exercise 7.2 Answers Question 2.
Show that Matrices And Determinants Class 11 Solutions Pdf Samacheer Kalvi Chapter 7 Ex 7.2
Solution:
Matrices And Determinants Class 11 State Board Solutions Samacheer Kalvi Chapter 7 Ex 7.2
11th Maths Matrices And Determinants Solutions Samacheer Kalvi Chapter 7 Ex 7.2

Matrices And Determinants Class 11 Solutions Pdf Question 3.
Prove that 11th Maths Exercise 7.2 In Tamil Samacheer Kalvi Chapter 7 Matrices And Determinants
Solution:
LHS
Taking a from C1, b from C2 and c from C3 we get
11th Maths Determinants Solutions Samacheer Kalvi Chapter 7 Ex 7.2
Expanding along R1 we get
(2c) (abc) (1) [ab + ab] = abc (2c) (2ab)
1 = (abc) (4abc) = 4a2b2c2
= RHS

Matrices And Determinants Class 11 State Board Solutions Question 4.
11th Maths Exercise 7.2 Samacheer Kalvi Chapter 7 Matrices And Determinants
Solution:
11th Maths Matrix Solutions Samacheer Kalvi Chapter 7 Matrices And Determinants Ex 7.2

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Prove that Exercise 7.2 Class 11 Maths Solutions State Board Samacheer Kalvi Chapter 7 Matrices And Determinants
Solution:
11th Maths Matrix And Determinants Samacheer Kalvi Chapter 7 Ex 7.2

11th Maths Exercise 7.2 In Tamil Question 6.
Show that 11th Maths Ex 7.2 Samacheer Kalvi Chapter 7 Matrices And Determinants
Solution:
Exercise 7.2 Class 11 Maths Solutions Chapter 7 Matrices And Determinants Samacheer Kalvi
11th Std Maths Determinants Solutions Samacheer Kalvi Chapter 7 Ex 7.2

11th Maths Determinants Solutions Question 7.
Write the general form of a 3 × 3 skew-symmetric matrix and prove that its determinant is 0.
Solution:
11th Maths Book Volume 2 Chapter 7 Samacheer Kalvi Matrices And Determinants Ex 7.2

11th Maths Exercise 7.2 Samacheer Kalvi Question 8.
Class 11 Maths Exercise 7.2 Solutions Samacheer Kalvi Chapter 7 Matrices And Determinants
Solution:
Matrices And Determinants Class 11 Solutions Samacheer Kalvi Chapter 7 Ex 7.2
we get – (aα2 + 2bα + c) [ac – b2]
So Δ = 0 ⇒ (aα2 + 2bα + c) (ac -b2) = – 0 = 0
⇒ aα2 + 2bα + c = 0 or ac – b2 = 0
(i.e.) a is a root of ax2 + 2bx + c = 0
or ac = b2
⇒ a, b, c are in G.P.

11th Maths Matrix Solutions Question 9.
Prove that Matrices And Determinants Class 11 Exercise Samacheer Kalvi Chapter 7 Ex 7.2
Solution:
Determinants Class 11 State Board Solutions Samacheer Kalvi Ex 7.2
11th Maths Matrices And Determinants Pdf Samacheer Kalvi Ex 7.2

Exercise 7.2 Class 11 Maths Solutions State Board Question 10.
If a, b, c are pth, qth and rth terms of an A.P., find the value of Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 21
Solution:
We are given a = tp,b = tq and c = tr
Let a be the first term and d be the common difference
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 22

11th Maths Matrix And Determinants Question 11.
Show that Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 23 is divisible by x4
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 24
Multiplying R1 by a, R2 by b and R3 by c and
taking out a from C1 b from C2 and c from C3 we get
=Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 25=

11th Maths Ex 7.2 Question 12.
If a, b, c are all positive, and are pth, qth and rth terms of a G.P., show that
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 26
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 27
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 28

Exercise 7.2 Class 11 Maths Solutions Question 13.
Find the value of Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 29 if x, y, z ≠ 1.
Solution:
Expanding the determinant along R1
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 30

11th Std Maths Determinants Solutions Question 14.
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 31
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 32
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 33

11th Maths Book Volume 2 Chapter 7 Question 15.
Without expanding, evaluate the following determinants:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 34
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 35

10th Maths Exercise 7.2 Samacheer Kalvi Question 16.
If A is a square matrix and |A| = 2, find the value of |AAT|.
Solution:
|A| = 2 (Given) |AT| = 2
Now |AAT| = |A| |AT| = 2 × 2 = 4.

12th Maths Exercise 7.2 Samacheer Kalvi Question 17.
If A and B are square matrices of order 3 such that |A| = -1 and |B| = 3, find the value of |3AB|.
Solution:
Given |A| = -1 : |B| = 3
Given A and B are square matrices of order 3.
∴ |kAB| = k3 |AB|
Here k = 3 ∴ |3AB| = 33 |AB|
= 27 |AB|
= 27 (-1) (3)
= -81

Class 11 Maths Exercise 7.2 Solutions Question 18.
If λ = -2, determine the value of Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 36
Solution:
Given λ = -2
∴ 2λ = -4; λ2 = (-2)2; 3λ2 + 1 = 3 (4) + 1 = 13
6λ – 1 = 6(-2) – 1 = -13
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 37
expanding along R1
0(0) + 4 (0 + 13) + 1 (-52 + 0) = 52 – 52 = 0
Aliter: The determinant value of a skew-symmetric matrix is zero

Matrices And Determinants Class 11 Solutions Question 19.
Determine the roots of the equation Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 38
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 39

Given the determinant value is 0
⇒ 30(1 + x) (2 – x) = 0
⇒ 1 + x = 0 or 2 – x = 0
⇒ x = -1 or x = 2
So, x = -1 or 2.

Matrices And Determinants Class 11 Exercise Question 20.
Verify that det (AB) = (det A) (det B) for Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 40
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 41
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 42
{(-20)(52) (-19) + (10)(38)(—49) + (2)(64)(-17)} – {(-49)(52) (2) + (-17)(38)(-20) + (-19)(64)(10)}
= (19760 – 18620 – 2176) – (-5096 + 12920 – 12160)
= (19760 + 5096 + 12160) – (18620 + 2176 + 12920)
= 37016 – 33716 = 3300 ….(3)
Now (1) × (2) = (3)
(i.e.,) (-33) (-100) = 3300
⇒ det (AB) = (det A), (det B)

Determinants Class 11 State Board Solutions Question 21.
Using cofactors of elements of second row, evaluate |A|, where Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 43
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 44

Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 Additional Problems

Exercise 7.2 Class 10 Samacheer Kalvi Question 1.
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 45
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 46
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 47

11th Maths Matrices And Determinants Pdf Question 2.
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 48
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 49

Question 3.
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 50
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 51
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 52

Question 4.
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 53
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 54
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 55
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 56

Question 5.
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 57
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 58

Question 6.
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 59
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 60
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 61

Question 7.
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 62
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 63

Question 8.
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 64
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 65
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 66

Question 9.
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 67
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 68
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 69

Samacheer Kalvi 11th Economics Solutions Chapter 1 Introduction To Micro-Economics

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Samacheer Kalvi 11th Economics Introduction To Micro-Economics Text Book Back Questions and Answers

Part – A
Multiple Choice Questions

Samacheer Kalvi 11th Economics Solution Chapter 1 Question 1.
‘Economics is a study of mankind in the ordinary business of life’ -It is the statement of
(a) Adam Smith
(b) Lionel Robbins
(c) Alfred Marshall
(d) Samuelson
Answer:
(c) Alfred Marshall

11th Economics Chapter 1 Book Back Answers Question 2.
The basic problem studied in Economics is ……………………….
(a) Unlimited wants
(b) Unlimited means
(c) Scarcity
(d) Strategy to meet all our wants
Answer:
(c) Scarcity

11th Economics 1st Lesson Questions And Answers Question 3.
Microeconomics is concerned with
(a) The economy as a whole
(b) Different sectors of an economy
(c) The study of individual economic units behaviour
(d) The interactions within the entir e economy
Answer:
(c) The study of individual economic units behaviour

Samacheer Kalvi 11th Economics Guide Question 4.
Which of the following is a microeconomics statement?
(a) The real domestic output increased by 2.5 percent last year
(b) Unemployment was 9.8 percent of the labor force last year
(c) The price of wheat determines its demand
(d) The general price level increased by 4 percent last year
Answer:
(c) The price of wheat determines its demand

11th Economics 1st Lesson Question 5.
Find the odd one out:
(a) “An inquiry into the nature and the causes of the Wealth of Nations”
(b) “Principles of Economics”
(c) “Nature and Significance of Economic Science”
(d) “Ceteris paribus”
Answer:
(d) “Ceteris paribus”

11th Economics Samacheer Kalvi Question 6.
The equilibrium price is the price at which ……………………….
(a) Everything is sold
(b) Buyers spend their money
(c) Quantity demanded equals quantity supplied
(d) Excess demand is zero
Answer:
(c) Quantity demanded equals quantity supplied

11th Economics Chapter 1 Question 7.
Author of “An Inquiry into the Nature and Causes of Wealth of Nations”
(a) Alfred Marshall
(b) Adam Smith
(c) Lionel Robbins
(d) Paul A Samuelson
Answer:
(d) Paul A Samuelson

Samacheer Kalvi Guru 11th Economics Question 8.
“Economics studies human behavior as a relationship between ends and scarce means which have alternative uses” is the definition of the economics of …………………………
(a) Lionel Robbins
(b) Adam Smith
(c) Alfred Marshall
(d) Paul A Samuelson
Answer:
(a) Lionel Robbins

Samacheer Kalvi 11th Economics Book Question 9.
Who is the Father of Economics?
(a) Max Muller
(b) Adam Smith
(c) Karl Marx
(d) Paul A Samuelson
Answer:
(b) Adam Smith

11th Economics Solutions Samacheer Kalvi Question 10.
“Economics is a science” The basis of this statement is …………………………
(a) Relation between cause and effect
(b) Use of deductive method and inductive method for the formations of laws
(c) Experiments
(d) All of the above
Answer:
(d) All of the above

11th Economics 1st Chapter Question Answer Question 11.
Utility means
(a) Equilibrium point at which demand and supply are equal
(b) Want-satisfying capacity of goods and services
(c) Total value of commodity
(d) Desire for goods and services
Answer:
(b) Want-satisfying capacity of goods and services

11th Economics Samacheer Kalvi Guide Question 12.
A market is ………………………
(a) Only a place to buy things
(b) Only a place to sell things
(c) Only a place where prices adjust
(d) A system where persons buy and sell goods directly or indirectly
Answer:
(d) A system where persons buy and sell goods directly or indirectly

Samacheer Kalvi 11th Economics Book Back Answers Question 13.
Which one of the following is not a point in the Welfare Definition of Economics?
(a) Study of an ordinary man
(b) Economics does not focus on wealth alone
(c) Economics is the study of material welfare
(d) Economics deals with unlimited wants and limited means
Answer:
(d) Economics deals with unlimited wants and limited means

11 Economics Samacheer Kalvi Question 14.
Growth definition takes into account ……………………….
(a) The problem of choice in the dynamic framework of Economics
(b) The problem of unlimited means in relation to wants
(c) The production and distribution of wealth
(d) The material welfare of human beings
Answer:
(a) The problem of choice in the dynamic framework of Economics

Samacheer Kalvi 11th Economics Question 15.
Which theory is generally included under microeconomics?
(a) Price Theory
(b) Income Theory
(c) Employment Theory
(d) Trade Theory
Answer:
(a) Price Theory

11th Economics 1st Chapter Question 16.
……………………….. have exchange value and their ownership rights can be established and exchanged.
(a) Goods
(b) Services
(c) Markets
(d) Revenue
Answer:
(a) Goods

11th Economics Chapter 1 Answers Question 17.
Identify the correct characteristics of utility
(a) It is equivalent to ‘usefulness’
(b) It has moral significance
(c) It is same as pleasure
(d) It depends upon consumer’s mental attitude
Answer:
(d) It depends upon consumer’s mental attitude

11 Th Samacheer Kalvi Economics Question 18.
Who has given a scarcity definition of economics?
(a) Adam Smith
(b) Marshall
(c) Robbins
(d) Robertson
Answer:
(c) Robbins

Economics Class 11 Samacheer Kalvi Question 19.
The process of reasoning from particular to general is
(a) Deductive method
(b) Inductive method
(c) Positive economics
(d) Normative economics
Answer:
(b) Inductive method

Question 20.
Total revenue is equal to total output sold multiplied by ………………………….
(a) Price
(b) Total cost
(c) Marginal revenue
(d) Marginal cost
Answer:
(a) Price

Part – B

Answer the following questions in one or two sentences

Economics Samacheer Kalvi Question 21.
What is meant by Economics?
Answer:
The term ‘Economics’ comes from oikonomikos which means ‘Management of households’. ‘Political Economy’ is renamed as ‘Economics’ by Alfred Marshall.

Economics Book Class 11 Samacheer Kalvi Question 22.
Define microeconomics?
Answer:

  1. Microeconomics is the study of the economic actions of individual units say households, firms, or industries.
  2. It studies how business firms operate under different market conditions.
  3. The combined actions of buyers and sellers determine prices.
    • Microeconomics covers:
    • Value theory [product pricing and factor pricing]
    • Theory of economic welfare.

Economics Class 11 Chapter 1 Questions And Answers Question 23.
What are the goods?
Answer:
The materialistic things and services which satisfy human wants are called as goods in economics.

Samacheer Kalvi Economics Question 24.
Distinguish goods from services.
Answer:

S.No. Goods Services
1. Material things, tangible Intangible
2. Exist independently of their owner Inseparable from their makers
3. Transferable have value – in – exchange Services cannot be stored as inventories like goods.

Question 25.
Name any two types of utility?
Answer:

  1. Time Utility: A sick man derives time utility from blood not at the time of its donation. but only at the operation – time, i.e., when it is used.
  2. Place Utility: A student derives place utility from a book not at the place of its publication (production centre) but only at the place of his education (consumption centre).

Question 26.
Define positive economics.
Answer:
Positive economics deals with what it is, its means, it analyses a problem on the basis of facts and examines its causes.

Question 27.
Give the meaning of the deductive method?
Answer:

  1. The deductive method is also named as an analytical or abstract method.
  2. It consists in deriving conclusions from general truths.
  3. It takes a few general principles and applies them to draw conclusions.
  4. The classical and neoclassical school of economists notably, Ricardo, JS Mill, Malthus Marshall, Pigou applied the deductive method in their economic investigations.

Part – C

Answer the following questions in One Paragraph

Question 28.
Explain the scarcity definition of Economics and assess it.
Answer:
Lionel Robbins published a book “An Essay on the Nature and Significance of Economic Science”. In it, he defined “Economics is a science which studies human behavior as a relationship between ends and scarce means which have alternative uses”.
Assessment:

  1. Robbins does not make any distinction between goods conducive to human welfare and goods that are not.
  2. Economics deals not only with the microeconomic aspects but also with the macroeconomic aspects like how national income is generated. Robbins reduces economics merely to the theory of resource allocation.
  3. Robbin’s definition does not cover the theory of economic growth and development.

Question 29.
What are the crucial decisions involved in ‘what to produce’?
Answer:
Every society much decides on what goods it will produce and how much of these it will produce.
In this process, the crucial decisions include:

  1. Whether to produce more of food, clothing and housing or to have more luxury goods.
  2. Whether to have more agricultural goods or to have industrial goods and services.
  3. Whether to use more resources in education and health or to use more resources in military services.
  4. Whether to have more consumption goods or to have investment goods.
  5. Whether to spend more on basic education or higher education.

Question 30.
Explain different types of economic activities.
Answer:

Samacheer Kalvi 11th Economics Solution Chapter 1 Introduction To Micro-Economics

  1. Economics focuses on the behavior and interactions among economic agents, individuals, and groups in the economic system.
  2. It deals with the consumption and production of goods and services and distribution of income among the factors of production.
  3. It includes the activities of the rational human beings under the existing social, legal, and institutional arrangement.
  4. It studies the way in which people use the available resources to satisfy their multiplicity of wants.

Question 31.
Elucidate different features of services?
Answer:
Along with goods, services are produced and consumed. They are generally, possess the following:

1. Intangible:

  • Intangible things are not physical objects but exist in connection to other things for example, brand image, goodwill etc.
  • The intangible things are converted and stored into tangible items such as recording a music piece into a pen – drive.

2. Heterogeneous:

  • Services vary across regions or cultural backgrounds.
  • A single type service yields multiple experiences, e.g., music, consulting physicians etc.

3. Inseparable from their makers:

  • Services are inextricably connected to their makes. For example, labour and labourers are inseparable.

4. Perishable:

  • Services cannot be stored as inventories like assets.
  • For example, it is useless to possess a ticket for a cricket-match once the match is over.
  • It cannot be stored and it has no value-in-exchange.

Question 32.
What are the important features of utility?
Answer:
Utility : Utility is the want satisfying power of a commodity or a service. Features of utility :

  1. Utility is psychological.
  2. Utility is not equivalent to usefulness.
  3. Utility is not the same as pleasure.
  4. Utility is personal and relative.
  5. Utility is the function of the intensity of human want.
  6. Utility is a subjective concept.
  7. Utility has no ethical or moral significance.

Question 33.
Distinguish between microeconomics and macroeconomics
Answer:

S.No. Micro Economics Macro Economics
1. It deals with the economic decision – making of individual economic agents. It deals with aggregates and averages of the entire economy.
2. It accounts only small components of the whole economy. It considers the economy of the country as a whole.
3. It deals with price determination of individual products and factors of pro­duction. It deals with general price – level in any economy.
4. It is concerned with the optimization goals of individual consumers and producers It is concerned with the optimization of the growth process of the entire economy.

Question 34.
Compare positive economics and normative economics.
Answer:
Positive economics deals with what it means, it analyses a problem on the basis of facts and examines its causes whereas, Normative economics responds to a question like what ought to be.

Positive Economics:

  1. An increase in the money supply implies a price rise in an economy.
  2. As the irrigation facilities and application of chemical fertilizers expand, the production of food – grains increases.
  3. An increase in birth rate and a decrease in the death rate reflect the rate of growth of the population.

Normative Economics:

  1. Inflation is better than deflation.
  2. More production of luxury goods in not good for a less developed country.
  3. Inequalities in the distribution of wealth and incomes should be reduced.

Part – D

Answer the following questions in about a page

Question 35.
Compare and contrast various definitions of Economics?
Answer:
1. Adam Smith – Wealth definition:

  1. Adam Smith [1723 – 1790], in his book “An Inquiry into Nature and Cause of Wealth of Nations” [1776] defines “Economics as the science of wealth”
  2. He explains how a nation’s wealth is created and increased.
  3. He considers that the individual in the society wants to promote his own gain and in this process, he is guided and led by an “invisible hand”
  4. Adam Smith favours the introduction of “division of labour” to increase the quantum of output.
  5. Severe competition in factories and society helps in bettering the product.
  6. Supply force is very active and a commodity is made available to the consumers at the lowest price.

2. Alfred Marshall – Welfare definition:

  1. Alfred Marshall [1842 – 1924] in his book “Principles of Economics” [1890] defines Economics thus “Political Economy” or Economics is a study of mankind in the ordinary business of life.
  2. It examines that part of individual and social action which is most closely connected with the attainment and with the use of the material requisites of well being.
  3. Thus, it is on one side a study of wealth and on the other and more important side, a part of the study of man”.

The important features of Marshall’s definition are:

  1. Economics does not treat wealth as the be all and end all of economic activities.
  2. Man promotes primarily welfare and not wealth. ‘
  3. The science of economics contains the concerns of ordinary people who are moved by love and not merely guided or directed by the desire to get maximum monetary benefit.
  4. Economics is a social science. It studies people in the society who influence one another.

3. Lionel Robbins – Scarcity definition:

  1. Lionel Robbins published a book “An Essay on the Nature and Significance of Economic Science” in 1932.
  2. According to him, “Economics is a science which studies human behaviour as a relationship between ends and scarce means which have alternative uses”.

The major features of Robbins’ definition:

  1. Ends refer to human wants. Human beings have unlimited number of wants.
  2. On the other hand, resources or means that go to satisfy the unlimited human wants are limited or scarce in supply.
  3. The scarce means are capable of having alternative uses.
  4. An individual grades his wants and satisfies first his most urgent want.
  5. Economics, according to Robbins, is a science of choice.

4. Samuelson’s – growth definition:
Paul Samuelson defines Economics as “the study of how men and society choose, with or without the use of money, to employ scarce productive resources which could have alternative uses to produce various commodities over time, and distribute them for consumption, now and in the future among various people and groups of society”.

The major implications of this definition are as follows:

  • Samuelson makes his definition dynamic by including the element of time in it.
  • Samuelson’s definition is applicable also in a barter economy.
  • His definition covers various aspects like production, distribution and consumption.
  • Samuelson treats Economics as a social science.
  • Samuelson appears to be the most satisfactory.

Question 36.
Explain various steps of deductive and inductive methods.
Answer:
Steps of deductive method :

  1. Should have clear and precise idea of the problem.
  2. Definition of technical terms and assumptions.
  3. Deducing hypothesis.
  4. Verification of hypotheses

Steps of inductive method :

  1. Data collection and arrangement.
  2. By observing the data conclusions are drawn easily.
  3. Generalization of data and hypothesis formulation.
  4. Verification of hypothesis.

Question 37.
Elaborate the nature and scope of Economics.
Answer:
Explanation :
Nature of Economics :
The nature of a subject refers to its contents and how and why they find a place in the subject. This nature is understood by studying the various definitions given by Economists.
The nature of Economics can be clearly understood from the following definitions :

  1. Adam Smith (classical Era) who considered Economics is a science of wealth gave wealth definition.
  2. Alfred Marshall (Neo-classical era) considered Economics as a social science which studies wealth on one side and the material welfare of human beings on the other side.
  3. Robbin’s Scarcity Definition (new age) He defined Economics is a science of choice.
  4. Samuelson gave growth definition which represents the modern age.
  5. The scope of economics refers to the subject matter of economics. It throws light on whether it is an art or a science and science, whether positive or normative science.

Economics: It’s the subject matter :

11th Economics Chapter 1 Book Back Answers Samacheer Kalvi Introduction To Micro-Economics

Related to socity:

  1. Economics focuses on the behaviour and interactions among economic agents, individuals and groups belonging to an economic system.
  2. It deals with the consumption and production of goods and services and the distribution of income among the factors of production.

Related to scarce resources :

  1. Economics studies the ways in which people use the available resources to satisfy their multiplicity of wants.

Human science or Social science :

  1. Economics is concerned with activities of human beings.
  2. The action of one member affect those of the others in the society. Hence, economics is called a human science or social science.

Related to wealth :

  1. Economics constitute all human activities related to wealth.
  2. Human activities not related to wealth are not included in Economics.

Economics as an art:

  1. Art is the practical application of knowledge for achieving particular goals.
  2. Economics provides guidance to the solutions to all the economic problems.

Economics as a science :

  1. Science is a systematic study of knowledge. Science develops the co-relation between cause and effect based on facts.
  2. Economics examines the relationship between the cause and effect of the problems. Hence, it is rightly considered as both an art and a science.

Economics : Positive and Normative science:

  1. Positive- Economics is concerned with how ? and why ? and normative Economics with ‘What ought to be’. Economics is both a positive and normative science.

Question 38.
Explain basic problems of the economy with the help of production possibility curve. Production possibility curve :
Answer:
The Problem of choice between relatively scarce commodities due to limited resources can be illustrated with the help of a geometric device, is known as production possibility curve

Basic problems of the Economy :

11th Economics 1st Lesson Questions And Answers Samacheer Kalvi Introduction To Micro-Economics

(i) The problem of choice :

  1. The problem of choice arise because of the given limited resources and unlimited wants.
  2. It may be related to the allocation of resources between people of different standards.
  3. Since PPC is the locus of the combination of the goods the problem of choice will not arise.

(ii) The Notion of scarcity :

  1. We can explain the notion of scarcity with the help of PPC.
  2. Every society possesses only a specific amount of resources, which can produce only limited amount of output even with the help of best technology.
  3. The PPC reflects the constraints imposed by the element of economic scarcity.

(iii) Solution of central problems :

  1. The central problems of an economy can be explained with the help of PPC.
  2. The solution for what to produce involves the decisionregarding the choice oflocation on the production possibility curves.
  3. A production combination represented by any point inside the PPC indicates that the economy is using inefficient methods of production and combination of resources.

Samacheer Kalvi 11th Economics Introduction To Micro-Economics Additional Questions and Answers

Part – A

Choose the best options

Question 1.
In Economics, we make use of ………………………
(a) Deductive Method
(b) Inductive Method
(c) Both
(d) None
Answer:
(c) Both

Question 2.
Adam Smith wrote “Wealth of Nations” in
(a) 1723
(b) 1890
(c) 1776
(d) 1932
Answer:
(c) 1776

Question 3.
Who wrote the book “Principles of Economics” in 1890?
(a) Adam Smith
(b) Alfred Marshall
(c) Lionel Robbins
(d) Samuelson
Answer:
(b) Alfred Marshall

Question 4.
Lionell Robbins book, “An essay on the nature and significance of economic science” was published in
(a) 1932
(b) 1776
(c) 1723
(d) 1890
Answer:
(a) 1932

Question 5.
Production refers to the creation of ……………………….
(a) Price
(b) Market
(c) Cost
(d) Utilities
Answer:
(d) Utilities

Question 6.
is the unit of measurement of utility.
(a) Price
(b) Utils
(c) Consumption
(d) Human wants
Answer:
(b) Utils

Question 7.
Perishable goods are ………………………..
(a) Long Term Lived
(b) Short Term Lived
(c) Medium Term Lived
(d) Longer Life Time Lived
Answer:
(b) Short Term Lived

Question 8.
“The proportion of total expenditure incurred on food items declines as total expenditure goes on increasing” is
(a) Marshall’s law
(b) Adam Smith’s law
(c) Engel’s law
(d) Samuelson’s law
Answer:
(c) Engel’s law

Question 9.
……………………… The economy is the other name for economics.
(a) Wealth
(b) Welfare
(c) Scarcity
(d) Political
Answer:
(d) Political

Question 10.
Utility can be measured indirectly using the ‘Measuring rod of money’ is the statement of
(a) Adam Smith
(b) Samuelson
(c) Marshall
(d) Lionel Robbins
Answer:
(c) Marshall

Question 11.
……………………. goods don’t directly satisfy the consumer.
(a) Capital
(b) Consumer
(c) Free
(d) Economic
Answer:
(a) Capital

Match the following and choose the answer using the codes given below

Question 1.

(a) Adam Smith (1) Nature and significance of economic science
(b) Alfred Marshall (2) Net economic welfare
(c) Robbins (3) Principles of Economics
(d) Samuelson (4) Wealth of nations

(a) 3 4 2 1
(b) 1 2 3 4
(c) 4 2 3 1
(d) 2 4 3 1
Answer:
(a) 3 4 2 1

Question 2.

(a) Goods (1) Man-made
(b) Services (2) Nature
(c) Free good (3) Tangible
(d) Consumer good (4) Intangible

(a) 1 2 3 4
(b) 2 4 3 1
(c) 4 3 2 1
(d) 3 4 2 4
Answer:
(c) 4 3 2 1

Choose the correct statement

Question 3.
(a) Utility is equal to usefulness
(b) Utility is same as pleasure
(c) Utility is a objective concept
(d) Utility has no ethical or moral significance
Answer:
(d) Utility has no ethical or moral significance

Question 4.
(a) Consumption is the result of co-orindation of factors of production
(b) Macro Economics is the obverse of micro Economics
(c) Macro Economics is called as price theory
(d) production studies about the pricing of factors of production
Answer:
(b) Macro Economics is the obverse of micro Economics

Choose the incorrect pair

Question 5.

(a) Micro Economics i Value theory
(b) Macro Economics ii Income theory
(c) International Economics iii Happiness index
(d) Public finance iv Financial Administration

Answer:
(c) International Economics – Happiness index

Question 6.

(a) Adam Smith i Classical era
(b) Marshall ii Neo classical era
(c) Robbins iii New age
(d) Samuelson iv Old age

Answer:
(d) Samuelson iv Old age

Choose the odd one out

Question 7.
(a) Form unity
(b) Place untility
(c) Time untility
(d) Total untility
Answer:
(d) Total untility

Question 8.
(a) Net economic
(b) Wealth defination
(c) welfare defination
(d) scarcity defination
Answer:
(a) Net economic

Choose the correct pair

Question 9.

(a) International economics i Public investment
(b) Health economics ii Drug price control
(c) Developmental economics iii Ecology and environment
(d) Environment economics iv Human development index

Answer:
(b) Health economics (ii) Drug price control

Question 10.

(a) Equilibrium i Keyres
(b) Utils ii Marshall
(c) Ragnar frisch iii England
(d) Neo-classical school iv Adam smith

Answer:
(b) Utils (ii) Marshall

Choose the incorrect statement

Question 11.
(a) Macro economics is concerned with the economy as a whole
(b) Micro economics is the study of economic actions of individual units
(c) Health economics is an area of applied economics
(d) International economics analyse the inter relationship between economy and environment.
Answer:
(d) International economics analyse the inter relationship between economy and environment.

Question 12.
(a) Economics is an art – A.C. Pigou, Marshall
(b) Economics is a science – Robbins, Jordan
(c) Economic laws are statement of tendencies – Adam Smith
(d) Political economy is said to have strangled itself with definitions – Keynes
Answer:
(c) Economic laws are statement of tendencies – Adam Smith

Analyse the reason for the following

Question 13.
Assertion (A) : Deductive method derives conclusions from general truth.
Reason (R) : Hypotheses can be verified through direct observation and statistical methods
(a) Both (A) and (R) are true, (R) is the correct explanation of (A)
(b) Both (A) and (R) are true, (R) is not the correct explanation of (A)
(c) (A) is true , (R) is false.
(d) Both (A) and (R) are false.
Answer:
(b) Both (A) and (R) are true, (R) is not the correct explanation of (A)

Question 14.
Assertion (A) : In every society certain choices have to be made.
Reason (R) : Resources are always scarce but wants are numerous.
(a) Both (A) and (R) are true, (R) is the correct explanation of (A)
(b) Both (A) and (R) are true, (R) is not the correct explanation of (A)
(c) (A) is true , (R) is false.
(d) (A) is false (R) is true.
Answer:
(a) Both (A) and (R) are true, (R) is the correct explanation of (A)

Fill in the blanks with the suitable option given below

Question 15.
The value of goods expressed in money terms is
(a) Revenue
(b) Market
(c) Price
(d) Cost
Answer:
(c) Price

Question 16.
Alfred Marshall wrote “Principles of Economics” in
(a) 1776
(b) 1932
(c) 1723
(d) 1890
Answer:
(d) 1890

Question 17.
______ is the assumption in economics that makes the economic laws hypothetical
(a) Other things remaining the same
(b) Time remaining the same
(c) Money remaining the same
(d) Wants to remain the same
Answer:
(a) Other things remaining the same

Choose the best option

Question 18.
Examples for perishable goods
(a) Television, furniture
(b) Table, chair
(c) Fish, fruits, flowers
(d) Vehicles, capital goods
Answer:
(c) Fish, fruits, flowers

Question 19.
The proportion of total expenditure incurred food items declines as total expenditure goes on increasing is
(a) Marshall’s law
(b) Samuelson’s law
(c) Adam smith’s law
(d) Engel’s law
Answer:
(d) Engel’s law

Question 20.
Utility can be measured indirectly using the ‘Measuring rod of money’ is the statement of
(a) Marshall
(b) Samuelson
(c) Robbins
(d) Frisch
Answer:
(a) Marshall

Part – B

Answer the following questions in one or two sentences

Question 1.
What is scarcity?
Answer:
Scarcity is the gap between what people want and what they are able to get.

Question 2.
Define normative economics?
Answer:

  1. Inflation is better than deflation.
  2. More production of luxury goods is not good for a less-developed country.
  3. Inequalities in the distribution of wealth and incomes should be reduced.

Question 3.
Explain Value.
Answer:
Value is the power of a commodity to command other commodities in exchange.

Question 4.
What is Income?
Answer:
Income is the amount of monetary or other returns, either earned or unearned, accruing over a period of time.

Question 5.
Give the meaning of the Inductive method.
Answer:
Induction is a process in logic facilitative or arriving at an inference, moving from particular to the general.

Question 6.
What are the kinds of goods?
Answer:

  1. Consumer goods
  2. Capital goods
  3. Perishable goods
  4. Durable goods

Part – C

Answer the following questions in One Paragraph

Question 1.
What are the four definitions of economics?
Answer:
Based on the particular stage of the growth of the subject of economics, the four definitions are presented here.

  1. Smith’s wealth definition, representing the classical era.
  2. Marshall’s welfare definition, representing the Neo-classical era.
  3. Robbin’s scarcity definition representing the new age.
  4. Samuelson’s growth definition representing the modern age.

Question 2.
Explain the steps of the deductive method?
Answer:
Steps of deductive method:

  • Step 1: The analyst must have a clear and precise idea of the problem to be inquired into.
  • Step 2: The analyst clearly defines the technical terms used in the analysis. Further, assumptions of the theory are to be precise.
  • Step 3: Deduce hypothesis from the assumptions taken.
  • Step 4: Hypotheses should be verified through direct observation of events in the real world and through statistical methods. [e.g.] There exists an inverse relationship between price and quantity demanded of a good.

Question 3.
Whether economics is an art or science ? Explain.
Answer:
1. Economics as an art:
Art is the practical application of knowledge for achieving particular goals. Economics provides guidance to the solutions to all the economic problems.
A.C. Pigou, Alfred Marshall and others regard economics as an art.

2. Economics as a science :
Science is a systematic study of knowledge. Science develops the co-relation between cause and effect based on facts.
Economics examines the relationship between the cause and effect of the problems. Hence, it is rightly considered as both an art and a science.

Question 4.
State the importance of microeconomics.
Answer:

  1. To understand the operation of an economy.
  2. To provide tools for economic policies.
  3. To examine the condition of economic welfare.
  4. Efficient utilization of resources.
  5. Useful in international trade.
  6. Useful in decision making.
  7. Optimal resource allocation.
  8. The basis for prediction.
  9. Price determination.

Question 5.
What are the methods of economic analysis ?
Answer:
There are two types of methods used in economics.
i. Deductive Method :

  1. It is also named as analytical or abstract method.
  2. It is a process in logic facilitating or arriving at an inference, moving from general to particular.
  3. It consists in deriving conclusions from general truths.

ii. Inductive method :
Inductive method is also called as empirical method. It involves the process of reasoning from particular facts to general principle. In this method, economic generalizations are derived based on

  1. Experimentations
  2. Observations
  3. Statistical methods

Alfred Marshall has rightly remarked “Inductive and Deductive methods are both needed for scientific thought as the right and left foot are both needed for walking”

Part – D

Answer the following questions in about a page

Question 1.
Explain the production possibility curve with a diagram.
The problem of choice between relatively scarce commodities can be illustrated with a help of a geometric device known as the production possibility curve.
Assumptions :
The analysis of the production possibility curve is based on certain assumptions :

  1. The time period remains constant.
  2. Techniques of production are fixed.
  3. Only two goods can be produced from the given resources.
  4. There is full employment in the economy.
  5. Resources of production are fully mobile.
  6. The factors of production are given in quantity and quality.
  7. The law of diminishing returns operates in production.

Production possibility schedule :

Production Possibilities Quantity of food production in tons No. of car produced
I 0 25
II 100 23
III 200 20
IV 300 15
V 400 8
VI 500 0

Explanation :
The above schedule suggests that if all resources are used for food production a maximum of 500 tons of food can be produced, given the existing technology. Instead, if all resources are used for producing cars, 25 cars can be produced. In between these two extreme possibilities exist, if we are willing to give up some food, we can have some cars.

Samacheer Kalvi 11th Economics Guide Solutions Chapter 1 Introduction To Micro-Economics

Diagram Explanation :

  • The quantity of food is shown on the x-axis and the number of cars on the y-axis.
  • Six different production possibilities P1, P2, P3, P4, P5, P6 are shown.
  • Point outside the curve P cannot be attained due to limited resources.
  • A point inside the curve P7 can be attained but at these points, resources are not fully employed

Question 2.
What are the basic economic problems?
Answer:
Basic economics problems:

  1. If resources are abundant and wants are so few, then there would be no economic problem.
  2. But this situation can never exist.
  3. Resources are always scarce and our wants are numerous.

Hence in every society certain choices have to be made.

The Economic Problem:

  1. Wants, desires; unlimited
  2. Resources: Scarce – Not freely available
  3. Economic choice
  4. Economics – How people use scarce resources to satisfy unlimited wants.

What and how much to produce?

  1. Every society must decide on what goods it will produce and how much of these it will produce.

In this process, the crucial decisions include:

  1. Whether to produce more of food, clothing and housing or to have more luxury goods.
  2. Whether to have more agricultural goods or to have industrial goods and services.
  3. Whether to use more resources in education and health or to use more resources in military services.
  4. Whether to have more consumption goods or to have investment goods.
  5. Whether to spend more on basic education or higher education.

How to produce?

  1. Every society has to decide whether it will use labor-intensive technology on capital intensive technology; that is whether to use more labour and less more machines and vice versa.

For whom to produce?

  1. Every society must also decide how its produce be distributed among the different sections of the society.
  2. It must also decide who gets more and who gets less.
  3. It should also decide whether or not a minimum amount of consumption be ensured for everyone in the society.
  4. Due to the scarcity of resources, a society faces the compulsion of making choice among alternatives.
  5. It face the problem of allocating the scare resources to the production of different possible goods and services and of distributing the produced goods and services among individuals within the economy.

If you need some more information about Tamilnadu State Board Solutions for 11th Economics Chapter 1 Introduction To Micro-Economics Questions and Answers then visit our site frequently and get all the resources that you look for. Also, share this page with your friends and make them learn all the concepts thoroughly before the exams.

Samacheer Kalvi 11th Accountancy Solutions Chapter 5 Trial Balance

Students can Download Accountancy Chapter 5 Trial Balance Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Accountancy Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Accountancy Solutions Chapter 5 Trial Balance

Samacheer Kalvi 11th Accountancy Trial Balance Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the Correct Answer

11th Accountancy Chapter 5 Book Back Answers Question 1.
Trial balance is a ………………
(a) Statement
(b) Account
(c) Ledger
(d) Journal
Answer:
(a) Statement

Trial Balance Questions With Solutions Pdf Question 2.
After the preparation of ledger, the next step is the preparation of ………………
(a) Trading account
(b) Trial balance
(c) Journal
(d) Profit and loss account
Answer:
(b) Trial balance

11th Accountancy 5th Chapter Solutions Question 3.
The trial balance contains the balances of ………………
(a) Only personal accounts
(b) Only real accounts
(c) Only nominal accounts
(d) All accounts
Answer:
(d) All accounts

Trial Balance Questions And Answers Pdf Question 4.
Which of the following is/are the objective(s) of preparing trial balance?
(a) Serving as the summary of all the ledger accounts
(b) Helping in the preparation of final accounts
(c) Examining arithmetical accuracy of accounts
(d) (a), (b) and (c)
Answer:
(d) (a), (b) and (c)

Accounting 11 Chapter 5 Answers Question 5.
While preparing the trial balance, the accountant finds that the total of the credit column is short by ₹ 200. This difference will be ………………
(a) Debited to suspense account
(b) Credited to suspense account
(c) Adjusted to any of the debit balance
(d) Adjusted to any of the credit balance
Answer:
(b) Credited to suspense account

Trial Balance Questions And Answers Question 6.
A list which contains balances of accounts to know whether the debit and credit balances are matched is ………………
(a) Journal
(b) Day book
(c) Trial balance
(d) Balance sheet
Answer:
(c) Trial balance

Trial Balance Question Question 7.
Which of the following method(s) can be used for preparing trial balance?
(a) Balance method
(b) Total method
(c) Total and Balance method
(d) (a), (b) and (c)
Answer:
(d) (a), (b) and (c)

Trial Balance Questions Question 8.
The account which has a debit balance and is shown in the debit column of the trial balance is ………………
(a) Sundry creditors account
(b) Bills payable account
(c) Drawings account
(d) Capital account
Answer:
(c) Drawings account

Trial Balance Question And Answer Question 9.
The difference of totals of both debit and credit side of trial balance is transferred to ………………
(a) Trading account
(b) Difference account
(c) Suspense account
(d) Miscellaneous account
Answer:
(c) Suspense account

Trial Balance Sums Question 10.
Trial balance is prepared ………………
(a) At the end of the year
(b) On a particular date
(c) For a year
(d) None of the above
Answer:
(b) On a particular date

II. Very Short Answer Questions

Trial Balance Class 11 Pdf Question 1.
What is trial balance?
Answer:
Trial balance is a statement containing the debit and credit balances of all ledger accounts on a particular date. It is arranged in the form of debit and credit columns placed side by side and prepared with the object of checking the arithmetical accuracy of entries made in the books of accounts and to facilitate preparation of financial statements.

Samacheer Kalvi Guru 11th Accountancy Question 2.
Give the format of trial balance.
Answer:
Trial balance is prepared in the following format under the balance method:
11th Accountancy Chapter 5 Book Back Answers Samacheer Kalvi Trial Balance

Trial Balance Questions With Solutions Question 3.
What are the methods of preparation of trial balance?
Answer:

  1. Balanced method
  2. Total method
  3. Total and balance method

Trial Balance Questions With Answers Question 4.
State whether the balance of the following accounts should be placed in the debit or the credit column of the trial balance:

  1. Carriage outwards
  2. Carriage inwards
  3. Sales
  4. Purchases
  5. Bad debts vi. Interest paid
  6. Interest received
  7. Discount received
  8. Capital
  9. Drawings
  10. Sales returns
  11. Purchase returns

Answer:
Trial balance as on 31st March, 2017
Trial Balance Questions With Solutions Pdf Samacheer Kalvi 11th Accountancy Solutions Chapter 5
III. Short Answer Questions

Trial Balance Chapter Pdf Question 1.
What are the objectives of preparing trial balance?
Answer:
Trial balance is prepared with the following objectives:
1. Test of arithmetical accuracy:
Trial balance is the means by which the arithmetical accuracy of the book – keeping work is checked. When the totals of debit column and credit column in the trial balance are equal, it is assumed that posting from subsidiary books, balancing of ledger accounts, etc. are arithmetically correct. However, there may be some errors which are not disclosed by trial balance.

2. Basis for preparing final accounts:
Financial statements, namely, trading and profit and loss account and balance sheet are prepared on the basis of summary of ledger balances obtained from the trial balance.

3. Location of errors:
When the trial balance does not tally, it is an indication that certain errors have occurred. The errors may have occurred at one or more of the stages of accounting process, namely, journalising or recording in subsidiary books, totalling subsidiary books, posting in ledger accounts.

Balancing the ledger accounts, carrying ledger account balances to the trial balance and totalling the trial balance columns, etc. Hence, the errors should be located and rectified before preparing the financial statements.

4. Summarised information of ledger accounts:
The summary of ledger accounts is shown in the trial balance. Ledger accounts have to be seen only when details are required in respect of an account.

Trial Balance Exercise With Answer Pdf Question 2.
What are the limitations of trial balance?
Answer:
The following are the limitations of trial balance:

  1. It is possible to prepare trial balance of an organisation, only if the double entry system is followed.
  2. Even if some transactions are omitted, the trial balance will tally.
  3. Trial balance may tally even though errors are committed in the books of account,
  4. If trial balance is not prepared in a systematic way, the final accounts prepared on the basis of trial balance may not depict the actual state of affairs of the concern.
  5. Agreement of trial balance is not a conclusive proof of arithmetical accuracy of entries made in the accounting records. This is because there are certain errors which are not : disclosed by trial balance such as complete omission of a transaction, compensating errors and error of principle.

Accountancy Class 11 Chapter 5 Solutions Question 3.
‘A trial balance is only a prima facie evidence of the arithmetical accuracy of records’. Do you agree with this statement? Give reasons.
Answer:
Yes, I agree with the statement. ‘A trial balance is only a prima facie evidence of the arithmetical accuracy of records.

Reasons:
Trial balance is the means by which the arithmetical accuracy of the book – keeping. work is checked. When the totals of debit column and credit column in the trial balance are equal, it is assumed that porting from subsidiary books, balancing of ledger accounts, etc.

IV. Exercises

Trial Balance Sums For Class 11 Question 1.
Prepare a trial balance with the following information: (3 Marks)
11th Accountancy 5th Chapter Solutions Samacheer Kalvi Trial Balance
Answer:
Trial balance
Trial Balance Questions And Answers Pdf Samacheer Kalvi 11th Accountancy Solutions Chapter 5

Chapter 5 Accounts Class 11 Solutions Question 2.
Prepare the trial balance from the following information: (3 Marks)
Accounting 11 Chapter 5 Answers Samacheer Kalvi Trial Balance
Answer:
Trial Balance
Trial Balance Questions And Answers Samacheer Kalvi 11th Accountancy Solutions Chapter 5

Samacheer Kalvi Accountancy 11th Question 3.
Prepare the trial balance from the following balances of Chandramohan as on 31st March, 2017. (3 Marks)
Trial Balance Question Samacheer Kalvi 11th Accountancy Solutions Chapter 5
Answer:
Trial balance of Chandramohan as on 31st March, 2017
Trial Balance Questions Samacheer Kalvi 11th Accountancy Solutions Chapter 5

Samacheer Kalvi 11th Accountancy Guide Question 4.
Prepare the trial balance from the following balances of Babu as on 31st March, 2016. (3 Marks)
Trial Balance Question And Answer Samacheer Kalvi 11th Accountancy Solutions Chapter 5
Answer:
Trial balance of Babu as on 31st March, 2016
Trial Balance Sums Samacheer Kalvi 11th Accountancy Solutions Chapter 5

Question 5.
From the following balances of Aijun, prepare the trial balance as on 31st March, 2018. (3 Marks)
Trial Balance Class 11 Pdf Samacheer Kalvi Chapter 5
Answer:
Trial balance of Arjun as on 31st March, 2018
Samacheer Kalvi Guru 11th Accountancy Solutions Chapter 5 Trial Balance

Question 6.
Prepare the trial balance from the following balances of Rajesh as on 31st March, 2017. (3 Marks)
Trial Balance Questions With Solutions Samacheer Kalvi 11th Accountancy Solutions Chapter 5
Answer:
Trial balance of Rajesh as on 31st March, 2017
Trial Balance Questions With Answers Samacheer Kalvi 11th Accountancy Solutions Chapter 5

Question 7.
Prepare the trial balance from the following balances of Karthik as on 31st March, 2017. (3 Marks)
Trial Balance Chapter Pdf Samacheer Kalvi 11th Accountancy Solutions Chapter 5
Answer:
Trial balance of Karthik as on 31st March, 2017
Trial Balance Exercise With Answer Pdf Samacheer Kalvi 11th Accountancy Solutions Chapter 5

Question 8.
From the following balance of Rohini, Prepare the trial balance as on 31st March, 2016. (3 Marks)
Accountancy Class 11 Chapter 5 Solutions Samacheer Kalvi Trial Balance
Answer:
Trial balance of Rohini as on 31st March, 2017
Trial Balance Sums For Class 11 Samacheer Kalvi Accountancy Solutions Chapter 5

Question 9.
Balan who has a car driving school gives you the following ledger balances. Prepare trial balance as on 31st December, 2016. (3 Marks)
Chapter 5 Accounts Class 11 Solutions Samacheer Kalvi Trial Balance
Answer:
Trial balance of Balan as on 31st March, 2017
Samacheer Kalvi Accountancy 11th Solutions Chapter 5 Trial Balance

Question 10.
The following balances are extracted from the books of Ravichandran on 31st December, 2016. Prepare the trial balance. (5 Marks)
Samacheer Kalvi 11th Accountancy Guide Chapter 5 Trial Balance
Answer:
Trial balance of Rohini as on 31st March, 2017
Samacheer Kalvi 11th Accountancy Solutions Chapter 5 Trial Balance

Question 11.
From the following balances, Prepare trial balanceof Baskar as on 31st March 2017. Transfer the difference, if any, to suspense account. (5 Marks)
Samacheer Kalvi 11th Accountancy Solutions Chapter 5 Trial Balance
Answer:
Trial balance of Rohini as on 31st March, 2017
Samacheer Kalvi 11th Accountancy Solutions Chapter 5 Trial Balance

Question 12.
From the following balances extracted from the books of Rajeshwari as on 31st March, 2017, prepare the trial balance.
Samacheer Kalvi 11th Accountancy Solutions Chapter 5 Trial Balance
Answer:
Trial balance of Rajeshwari as on 31st March, 2017
Samacheer Kalvi 11th Accountancy Solutions Chapter 5 Trial Balance

Question 13.
Correct the following trail balance:
Samacheer Kalvi 11th Accountancy Solutions Chapter 5 Trial Balance
Answer:
The Corrected Trial balance
Samacheer Kalvi 11th Accountancy Solutions Chapter 5 Trial Balance

Textbook Case Study Solved

Question 1.
Mary runs a textile store. She has prepared the following trial balance from her ledger balances. Her trial balance does not tally. She needs your help to check whether what she has done is correct.
Samacheer Kalvi 11th Accountancy Solutions Chapter 5 Trial Balance
Solution:
There are some errors in Mary’s Trial balance. The corrected trial balance has been written below:
The Corrected Trial balance of Mary
Samacheer Kalvi 11th Accountancy Solutions Chapter 5 Trial Balance

Samacheer Kalvi 11th English Reading Note-Making and Summarizing

Students who are interested in learning of 11th English Reading Note-Making and Summarizing Questions and Answers can use Tamilnadu State Board Solutions of 11th English Chapter Wise Pdf. First check in which chapter you are lagging and then Download Samacheer Kalvi 11th English Book Solutions Questions and Answers Summary, Activity, Notes Chapter Wise. Students can build self confidence by solving the solutions with the help of Tamilnadu State Board English Solutions. English is the scoring subject if you improve your grammar skills. Because most of the students will lose marks by writing grammar mistakes. So, we suggest you to Download Tamilnadu State Board 11th English Solutions according to the chapters.

Tamilnadu Samacheer Kalvi 11th English Reading Note-Making and Summarizing

Check out the topics covered in Reading Note-Making and Summarizing Questions and Answers before you start your preparation. Improve your grammar skills with the help of Samacheer Kalvi 11th English Book Solutions Questions and Answers pdf links. The solutions for Tamilnadu State Board 11th English Textbook are prepared by the English experts. So, if you follow Tamilnadu State Board Solutions 11th English Textbook Solutions you can cover all the topics in Reading Note-Making and Summarizing Questons and Answers. This helps to improve your communication skills.

Notes are short written record of facts to aid the memory. Notes are usually taken to record a speech or dictation while listening to it or after reading a book, magazine or article. They are referred back whenever needed and may be reproduced in the desired way.

 

A Good Business Letter Note Making The Necessity Of Note-Making

Knowledge is vast and unlimited, but our memory is limited. We cannot remember all the information all the time. Hence note-making is necessary. With the help of notes we can recall the entire information read/heard months ago. It is quite useful to students preparing for many subjects. At the time of examinations, it is not possible to go through voluminous books. At such critical times, notes are quite handy. Hence note-making fulfils three useful functions:
(i) It keeps a lot of information at our disposal for ready reference.
(ii) It helps us reconstruct what was said or written and thus accelerates the process of remembering/recall.
(iii) It comes in handy in delivering a speech, participation in a debate/discussion, writing an essay and revising lessons before an examination.

How Note-Making Helps Us

While making notes we do not simply read the passage/listen to speech but consider various points made by the writer/speaker and draw our own inferences about what is being presented. Thus note-making helps us in understanding the passage in a better way and organising our thoughts systematically.

 

Characteristics Of Good Notes

  • Short and Compact: Good notes must be short and compact.
  • Complete Information: They must contain all the important information.
  • Logical: They must be presented in a logical way.
  • Understandable: They should be understandable when consulted at a later stage.

Mechanics Of Note-Making

While making notes we follow certain standard practices. These may be listed as follows:
(a) Heading and Sub-headings
(b) Abbreviation and Symbols
(c) Note-form
(d) Numbering and Indentation

Heading And Sub-Headings
The heading reflects the main theme whereas the sub-headings point out how it has been developed. The selection of proper heading and sub-heading reveals the grasp of the passage by the students. In the absence of proper assimilation of main ideas and subsidiary points it is impossible to make notes.

 

Abbreviations And Symbols
They are used for precision and economy of words and hence quite helpful in note-making. At least four recognisable abbreviations are to be used in note-making in your board examination. These are essential components of note-making. Students often make use of abbreviations and symbols in doing their written work.

Conversation Is Indeed The Most Easily Solved Questions
Read the following passages carefully:

Conversation Is Indeed The Most Easily Teachable Passage Answers Passage 1
1. The conversation is indeed the most easily teachable of all arts. All you need to do in order to become a good conversationalist is to find a subject that interests you and your listeners. There are, for example, numberless hobbies to talk about. But the important thing is that you must talk about other fellow’s hobby rather than your own. Therein lies the secret of your popularity. Talk to your friends about the things that interest them, and you will get a reputation for good fellowship, charming wit, and a brilliant mind. There is nothing that pleases people so much as your interest in their interest.

 

2. It is just as important to know what subjects to avoid and what subjects to select for good conversation. If you don’t want to be set down as a wet blanket or a bore, be careful to avoid certain unpleasant subjects. Avoid talking about yourself, unless you are asked to do so. People are interested in their own problems, not in yours. Sickness or death bores everybody. The only one who willingly listens to such talk is the doctor, but he gets paid for it.

3. To be a good conversationalist you must know not only what to say, but how also to say it. Be mentally quick and witty. But don’t hurt others with your wit. Finally, try to avoid mannerisms in your conversation. Don’t bite your lips or click your tongue, or roll your eyes or use your hands excessively as you speak.

4. Don’t be like that Frenchman who said, “How can I talk if you hold my hand?”

11th English Summary Writing Questions:
A. On the basis of your understanding of the above passage make notes on it using headings and sub-headings. Use recognizable abbreviations wherever necessary. Give an appropriate title.
B. Write a summary of the above passage in about 80 words.
Answers:
A. TITLE: The Art of Conversation NOTES:
1. Conv’n—most easily tch’ble art
(i) Reqd. interest’g subject – hobbies
(ii) Talk abt other fellow’s int./hobby
(iii) Win’g reptn. as good conversationalist
(a) good f ’ship
(b) charm’g wit
(c) brl. mind

 

2. Fit subs, for conversationalist
(i) What subs, to avoid/select?
(ii) Avoid unpl’nt subs.
(a) sickness
(b) death
(iii) Avoid talk’g abt self

3. Qualities A a good conversationalist
(i) What to say & how to say it
(ii) ment’y quick & witty
(iii) pleasant & unhurt’g
(iv) avoid mannerisms

Key to Abbreviations and Symbols used

  • Conv’n – Conversation
  • tch’ble – teachable
  • Reqd. – Required
  • interest’g – interesting
  • abt – about
  • int. – interest
  • Win’g – Winning
  • reptn. – reputation
  • f’ship – friendship
  • charm’g – charming
  • brl. – brilliant
  • subs. – subjects
  • unpl’nt – unpleasant
  • talk’g – talking
  • A – of
  • ment’y – mentally
  • & – and
  • unhurt’g – unhurting

 

B. Summary
Conversation is the easiest and the most effective tool than other arts. To have such attractive quality, you need to pick a subject that interests your listeners more than you. Talk to your friends on topics that can indulge your friends in the conversation for a longer period of time. Being a good conversationalist, you have to be quick and witty. You should have a pleasant and unhurting quality. Mannerism should be avoided.

Passage 2
1. A good business letter is one that gets results. The best way to get results is to develop a letter that, in its appearance, style and content, conveys information efficiently. To perform this function, a business letter should be concise, clear and courteous.

2. The business letter must be concise: don’t waste words. Little introduction or preliminary chat is necessary. Get to the point, make the point, and leave it. It is safe to assume that your letter is being read by a very busy person with all kinds of papers to deal with. Re-read and revise your message until the words and sentences you have used are precise. This takes time, but is a necessary part of a good business letter. A short business letter that makes its point quickly has much more impact on a reader than a long-winded, rambling exercise in creative writing. This does not mean that there is no place for style and even, on occasion, humour in the business letter. While it conveys a message in its contents, the letter also provides the reader with an impression of you, its author: the medium is part of the message.

 

3. The business letter must be clear. You should have a very firm idea of what you want to say, and you should let the reader know it. Use the structure of the letter—the paragraphs, topic sentences, introduction and conclusion—to guide the reader point by point from your thesis, through your reasoning, to your conclusion. Paragraph often, to break up the page and to lend an air of organisation to the letter. Use an accepted business-letter format. Re-read what you have written from the point of view of someone who is seeing it for the first time, and be sure that all explanations are adequate, all information provided (including reference numbers, dates, and other identification). A clear message, clearly delivered, is the essence of business communication.

4. The business letter must be courteous. Sarcasm and insults are ineffective and can often work against you. If you are sure you are right, point that out as politely as possible. Explain why you are right, and outline what the reader is expected to do about it. Another form of courtesy j is taking care in your writing and typing of the business letter. Grammatical and spelling errors (even if you call them typing errors) tell a reader that you don’t think enough of him or can > lower the reader’s opinion of your personality faster than anything you say, no matter how idiotic. There are excuses for ignorance; there are no excuses for sloppiness.

 

5. The business letter is your custom-made representative. It speaks for you and is a permanent record of your message. It can pay big dividends on the time you invest in giving it a concise message, a clear structure, and a courteous tone.

Note Making Conversation Is Indeed Questions:
A. On the basis of your understanding of the above passage make notes on it using headings and sub-headings. Use recognizable abbreviations wherever necessary. Give an appropriate title.
B. Write a summary of the above passage in about 80 words.
Answers:
A. TITLE: Writing a Business Letter
Notes:
1. Features of a gd. busns. letter
(i) conveys info efficiently to get results
(ii) is concise
(iii) is clear
(iv) is courteous

 

2. How to write a gd. busns. letter
(i) Making letter concise
(a) Intro shd be brief
(b) make ur pt in precise words and sent’s
(c) short letr more effective
(d) style is imp.—may ocasnly have hum’
(ii) Achieving clarity
(a) Have a clear idea of what you wish to say
(b) structr the letter—intro & conclsn.
(c) use accepted format; para, topic, sent’s
(d) check facts, expl’ns, refs.
(iii) Bejng courteous
(a) Expin. ur pt. politely—avoid sarcasm/insults.
(b) careful wrtg & typg.
(c) gram. & spel’g errors to be avoided

3. Importance of busns. letr
(i) a representative
(ii) permanent rec. message

 

Key to Abbreviations and Symbols used

  • gd – good
  • busns – business
  • info – information
  • shd – should
  • letr – letter
  • pt – point
  • sent’s – sentences
  • ur – your
  • imp – important
  • ocasnly – occasionally
  • hum’r – humour
  • & – and
  • structr – structure
  • intro – introduction
  • conclsn – conclusion
  • para – paragraphs
  • expl’ns – explanations
  • refs – references
  • Expln – Explain
  • wrtg – writing
  • typg – typing
  • gram – grammar
  • spel’g – spelling
  • rec – record

 

B. Summary
A good business letter is that lends you positive and quality results. To get such results, a business letter should be effective in appearance, style and content. Apart from this a letter should be concise, clear and courteous. The business letter should be to the point as the message can be clear to the reader with an impression of you. The structure of letter should have topic sentence, introduction, paragraphs to conclusion. Re-read the points you have written to avoid sarcasm and insults that can work against your motive. Further more grammar and spelling errors need to be avoided.

The main aim is to share the knowledge and help the students of 11th English to secure the best score in their final exams. Use the concepts of Samacheer Kalvi 11th English Book Solutions Reading Note-Making and Summarizing Questions and Answers in Real time to enhance your skills. If you have any doubts you can post your comments in the comment section, We will clarify your doubts as soon as possible without any delay.

Samacheer Kalvi 11th Computer Science Solutions Chapter 5 Working with Typical Operating System (Windows & Linux)

Students can Download Computer Science Chapter 5 Working with Typical Operating System (Windows & Linux) Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Computer Science Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Computer Science Solutions Chapter 5 Working with Typical Operating System (Windows & Linux)

Samacheer Kalvi 11th Computer Science Working with Typical Operating System (Windows & Linux) Text Book Back Questions and Answers

PART – 1
I. Choose The Correct Answer

11th Computer Science Chapter 5 Book Back Answers Question 1.
From the options given below, choose the operations managed by the operating system …………….
(a) Memory
(b) Processor
(c) I/O devices
(d) all of the above
Answer:
(d) all of the above

11th Computer Science 5th Lesson Book Back Answers Question 2.
Which is the default folder for many Windows Applications to save your file?
(a) My Document
(b) My Pictures
(c) Documents and Settings
(d) My Computer
Answer:
(a) My Document

Chapter 5 Computer Science Class 11 Question 3.
Under which of the following OS, the option Shift + Delete – permanently deletes a file or folder?
(a) Windows 7
(b) Windows 8
(c) Windows 10
(d) all of the OS
Answer:
(d) all of the OS

Samacheer Kalvi 11th Computer Application Question 4.
What is the meaning of “Hibernate” in Windows XP/Windows 7?
(a) Restart the Computer in safe mode
(b) Restart the Computer in hibernate mode
(c) Shutdown the Computer terminating all the running applications
(d) Shutdown the Computer without closing the running applications
Answer:
(c) Shutdown the Computer terminating all the running applications

11th Computer Application Samacheer Kalvi Question 5.
Which of the following OS is not based on Linux?
(a) Ubuntu
(b) Redhat
(c) Cent OS
(d) BSD
Answer:
(d) BSD

Question 6.
Which of the following in Ubuntu OS is used to view the options for the devices installed?
(a) Settings
(b) Files
(c) Dash
(d) VBox_GAs_5.2.2
Answer:
(d) VBox_GAs_5.2.2

Question 7.
Identify the default email client in Ubuntu …………….
(a) Thunderbird
(b) Firefox
(c) Internet Explorer
(d) Chrome
Answer:
(a) Thunderbird

Question 8.
Which is the default application for spreadsheets in Ubuntu? This is available in the software launcher ……………..
(a) LibreOffice Writer
(b) LibreOffice Calc
(c) LibreOffice Impress
(d) LibreOffice Spreadsheet
Answer:
(b) LibreOffice Calc

Question 9.
Which is the default browser for Ubuntu?
(a) Firefox
(b) Internet Explorer
(c) Chrome
(d) Thunderbird
Answer:
(a) Firefox

Question 10.
Where will you select the option to log out, suspend, restart, or shut down from the desktop of Ubuntu OS?
(a) Session Indicator
(b) Launcher
(c) Files
(d) Search
Answer:
(a) Session Indicator

PART – 2
II. Short Answers

Question 1.
Differentiate cut and copy options.
Answer:
Cut:

  • The file is moved to the destination location.
  • Edit → cut, Edit → paste is used for cut.
  • Ctrl + X, Ctrl + V is used.
  • Right click ? cut, Right click ? paste is used.

Copy:

  • File will be present in both source destination places.
  • Edit → copy, Edit → paste is used for copy.
  • Ctrl + C, Ctrl + V is used.
  • Right click → copy, Right click → paste is used.

Question 2.
What is the use of a file extension?
Answer:
File Extension is the second part of the file name. It succeeds the decimal point in the file name. It is used to identify the type of file and it is normally upto 3 to 4 characters long.
Example: exe, html.

Question 3.
Differentiate Files and Folders.
Answer:
The basic difference between the two is that files store data, while folders store files and other folders.
Files:
Files, on the other hand, can range from a few bytes to several gigabytes. They can be documents, programs, libraries, and other compilations of data.

Folders:
The folders, often referred to as directories, are used to organize files on the computer. The folders themselves take up virtually no space on the hard drive.

Question 4.
Differentiate Save and save As option.
Answer:
Save:
This will save the document without asking for a new name or location. It will over – write the original.

Save as:
This will prompt to save the document using a dialog box. You will have the ability to change the file name and/or location. If you choose the same file name and location it will over – write the original. Your working document will be the one you just saved.

Question 5.
What is Open Source?
Answer:
Open source refers to a program or software in which the source code is available in the web to all public, without any cost.

Question 6.
What are the advantages of open source?
Answer:
Advantages:

  1. It is free of cost
  2. Accessible to everybody.
  3. Not belongs to particular vendor.
  4. It can be modified, for requirements.

Question 7.
Mention the different server distributions in Linux OS.
Answer:
The server distribution in Linux OS:

  1. Ubuntu Linux
  2. Linux Mint
  3. Arch Linux
  4. Deepin
  5. Fedora
  6. Debian
  7. Centos

Question 8.
How will you log off from Ubuntu OS?
Answer:
By choosing an option logout through the session indicator on the far right side of the top panel.

PART – 3
III. Explain in Brief

Question 1.
Analyse: Why the drives are segregated?
Answer:

  1. It saves space and increases system performance.
  2. Sharing and protection.
  3. Segment and segregate the data to defend it from cyber attack.

Question 2.
If you are working on multiple files at a time, sometimes the system may hang.
What is the reason behind it. How can you reduce it?
Answer:
When we open multiple applications, each application opened on the system takes some internal and hardware resources to keep it running. Memory is used by all the applications. System processing speed will be shared and it will become slow finally the system will hang.

Methods of reducing it:

  1. It is advisable to run one program at a time.
  2. Upgrade RAM
  3. Clean the cache Memories
  4. Defragmentation can be done

Question 3.
Are drives such as hard drive and floppy drives represented with drive letters? If so why, if not why?
Answer:
Yes, hard drives and floppy drives are represented with drive letters.
A: drive is used for floppy disk of 3.5 inches and storage capacity of 1.44 MB.
B: drive is for floppy of size 5.25 inches and of storage capacity of 1.2 MB
So, we can say like
A: First Floppy Drive
B: Second Floppy Drive
C: D : E: …………….. Z: Hard Disk Drives, CD/DVD

Question 4.
Write the specific use of Cortana.
Answer:
Uses of Cortana:

  1. Cortana is a voice activated personal assistant.
  2. Give the reminders based on time, places, or people.
  3. Track packages, teams, interests, and flights.
  4. Send emails and texts.
  5. Manage the calendar and keeps up to date.
  6. Create and manage lists.
  7. Chit chat and play games.
  8. Find facts, files, places, and info.
  9. Open any app on the system.

Question 5.
List out the major differences between Windows and Ubuntu OS.
Answer:

  1. Ubuntu supports high security than windows.
  2. Windows is more user friendly than Ubuntu
  3. Ubuntu is a free and open source based Linux OS whereas windows is a closed source and it is not free.
  4. Ubuntu uses Linux kernel whereas windows uses a hybrid kernel.

Question 6.
Are there any difficulties you face while using Ubuntu? If so, mention it with reasons.
Answer:

  1. Ubuntu has less hardware support for commercial / industrial / medical/ logistical therefore its less votted for use in big time backends.
  2. Ubuntu doesn’t support middlewares such as C panel, cloud linux and a plethora of other infrastructure or monitoring tools.
  3. Hard to install graphic drivers especially for old hardwares. It is not possible to play the modern games, because of poor graphics quality.
  4. The user switching from windows will not like the user experience on Ubuntu and will have difficulty in operating the OS.
  5. Ubuntu is not capable of playing MP3 files by default.

Question 7.
Differentiate Thunderbird and Firefox in Ubuntu OS.
Answer:
Difference between Thunderbird and Firefox in Ubuntu:

  1. Thunderbird is inbuilt email software of ubuntu whereas Firefox is a web browser.
  2. Thunderbird is used to access email such as Gmail, hotmail whereas Firefox is used to open web pages including e-mail.

Question 8.
Differentiate Save, Save As and Save a Copy in Ubuntu OS.
Answer:
11th Computer Science Chapter 5 Book Back Answers Samacheer Kalvi Working With Typical Operating System

PART – 4
IV. Explain in Detail

Question 1.
Explain the versions of Windows Operating System.
Answer:
11th Computer Science 5th Lesson Book Back Answers Samacheer Kalvi Working With Typical Operating System

Question 2.
Draw and compare the icon equivalence in Windows and Ubuntu.
Answer:
Chapter 5 Computer Science Class 11 Samacheer Kalvi Working With Typical Operating System
Elements of Ubuntu:
Search your Computer Icon : This icon is equal to search button in Window’s OS. Here, you have to give the name of the File or Folder for searching them.

Files : This icon is equivalent to My Computer icon. From here, you can directly go to Desktop, Documents and so on.

Firefox Web Browser : By clicking this icon, you can directly browse the internet. This is equivalent to clicking the Web Browser in Task bar in Windows.

Libre Office Writer : This icon will directly take you to document preparation application like MS Word in Windows.

Libre Office Calc: This icon will open LibreOffice Calc application. It is similar to MS Excel in Windows.

LibreOffice Impress : By clicking this icon, you can open LibreOffice Impress to prepare any presentations in Ubuntu like MS PowerPoint.

Ubuntu Software Icon : This icon will let you add any additional applications you want. This can be done by clicking the Update option at the top right comer of that screen.

Online Shopping icon : Using this icon user can buy and sell any products online.

System Settings Icons : This icon is similar to the Control panel in the Windows Operating System. But here, you need to authenticate the changes by giving your password. You cannot simply change as you do in Windows.

Trash : This icon is the equivalent of Recycle bin of windows OS. All the deleted Files and Folders are moved here.

Question 3.
Complete the following matrix.
Samacheer Kalvi 11th Computer Application Science Solutions Chapter 5 Working With Typical Operating System
Answer:
11th Computer Application Samacheer Kalvi Science Solutions Chapter 5 Working With Typical Operating System

Question 4.
Observe the figure and mark all the window elements. Identify the version of the Windows OS.
Answer:
Samacheer Kalvi 11th Computer Science Solutions Chapter 5 Working with Typical Operating System (Windows & Linux) 1

Question 5.
Write the procedure to create, rename, delete and save a file in Ubuntu OS. Compare it with Windows OS.
Answer:
Samacheer Kalvi 11th Computer Science Solutions Chapter 5 Working with Typical Operating System (Windows & Linux)

Samacheer kalvi 11th Computer Science Working with Typical Operating System (Windows & Linux) Additional Questions and Answers

PART – 1
I. Choose the correct answer

Question 1.
………………. is Open source Operating System for desktop and server.
(a) Windows series
(b) Android
(c) iOS
(d) Linux
Answer:
(d) Linux

Question 2.
The most common way of opening a file or a Folder is to click on it.
(a) left
(b) right
(c) double
(d) single
Answer:
(c) double

Question 3.
If you want to select multiple files or folders, use ……………….
(a) Ctrl + shift
(b) Ctrl + click
(c) shift + click
(d) Ctrl + shift + click
Answer:
(b) Ctrl + click

Question 4.
………………. is a special folder to keep the files or folders deleted by the user, which means you still have an opportunity to recover them.
(a) My computer
(b) Documents
(c) Recycle bin
(d) Pictures
Answer:
(c) Recycle bin

Question 5.
………………. is one of the popular Open Source versions of the UNIX Operating System.
(a) Windows 7
(b) Windows 8
(c) Linux
(d) Android
Answer:
(c) Linux

Question 6.
………………. icon is equivalent to My Computer icon. From here, you can directly go to Desktop, Documents and so on.
(a) Files
(b) Documents
(c) Downloads
(d) Computer
Answer:
(a) Files

Question 7.
………………. icon is the equivalent of Recycle bin of windows OS. All the deleted Files and Folders are moved here.
(a) Trash
(b) Files
(c) Online shopping
(d) Libre Office Impress
Answer:
(a) Trash

Question 8.
The vertical bar of icons on the left side of the desktop is called the ………………..
(a) Search
(b) Libre office calc
(c) Launcher
(d) Files
Answer:
(c) Launcher

Question 9.
………………. manages network connections, allowing you to connect to a wired or wireless network.
(a) Toolbar
(b) Title bar
(c) Session indicator
(d) Network indicator
Answer:
(d) Network indicator

Question 10.
To permanently delete a file or folder (i.e. to avoid sending a file or folder to the Recycle Bin), hold down the SHIFT key, and press on the keyboard.
(a) restore
(b) delete
(c) send to
(d) cut
Answer:
(b) delete

Question 11.
Clock is available in ……………….
(a) system tray
(b) Files
(c) start
(d) My documents
Answer:
(a) system tray

Question 12.
………………. command should be typed in the Run dialog bar to open Calculator?
(a) Calculator
(b) Calc
(c) Arithmetic
(d) Calculator open
Answer:
(b) Calc

Question 13.
The menu bar is present below the ……………….
(a) Task bar
(b) Scroll bar
(c) Title bar
(d) Function bar
Answer:
(c) Title bar

Question 14.
Which of the following OS bar plug and play feature ……………….
(a) Window XP
(b) Windows 98
(c) Windows 95
(d) Windows me
Answer:
(b) Windows 98

Question 15.
………………. has the task for frequently used applications?
(a) Quick Launch Tool bar
(b) Settings
(c) My pc
(d) This pc
Answer:
(a) Quick Launch Tool bar

Question 16.
The winkey combination used to display desktop is ……………….
(a) winkey + dt
(b) winkey + T
(c) winkey + alt + D
(d) winkey + D
Answer:
(d) winkey + D

Question 17.
SSD stands for ……………….
(a) Solid State Devices
(b) Simple Stage Driver
(c) Single State Drivers
(d) Synchronized State Devices
Answer:
(a) Solid State Devices

Question 18.
The mouse pointer becomes ………………. when it is positioned over a border or a comer of a window.
(a) +
(b) arrow
(c) single headed arrow
(d) double headed arrow
Answer:
(d) double headed arrow

Question 19.
What is the name given to the document window to enter or type the text?
(a) Work space
(b) Work Area
(c) Typing Area
(d) Space
Answer:
(a) Work space

Question 20.
………………. enables alternate method of opening an application.
(a) Running
(b) Searching
(c) Run
(d) Open
Answer:
(c) Run

Question 21.
The disk drives mounted in the system can be seen by clicking ……………….
(a) Disk drive Icon
(b) Drive Icon
(c) Device Driver Icon
(d) My Computer Icon
Answer:
(d) My Computer Icon

Question 22.
What is the name given to the rectangular area in an application or a document?
(a) Document
(b) Window
(c) Application
(d) Desktop
Answer:
(b) Window

Question 23.
Windows 10 was developed in the year ……………….
(a) 2009
(b) 2012
(c) 2015
(d) 2018
Answer:
(c) 2015

Question 24.
Match the following:
Samacheer Kalvi 11th Computer Science Solutions Chapter 5 Working with Typical Operating System (Windows & Linux)
(a) (3) (4) (2) (1)
(b) (1) (2) (3) (4)
(c) (4) (3) (2) (1)
(d) (4) (2) (1) (3)
Answer:
(a) (3) (4) (2) (1)

Question 25.
The Rulers are used to set ……………….
(a) Orientations
(b) Header
(c) Footer
(d) Margins
Answer:
(d) Margins

Question 26.
Which one of the following boots faster, mns apps faster compared to HDD.
(a) FDD
(b) Cache
(c) SSD
(d) DVD
Answer:
(c) SSD

Question 27.
Which functional key is used to bring the focus on, the first menu of the menu bar?
(a) F5
(b) F10
(c) F11
(d) F7
Answer:
(b) F10

Question 28.
How many disk drive icon options are there?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(d) 5

Question 29.
Which one of the following is used to open search results dialog box?
(a) search
(b) See more results
(c) search more results
(d) searching web
Answer:
(b) See more results

Question 30.
Which icon is used to check whether one system is connected to another system?
(a) Network
(b) System
(c) Control panel
(d) Hard drive
Answer:
(a) Network

Question 31.
The keyboard shortcut to save a file is ……………….
(a) alt + s
(b) Ctrl + s
(c) Ctrl + alt + s
(d) winkey + s
Answer:
(b) Ctrl + s

Question 32.
Which command is used to create new folder?
(a) File → folder
(b) File → New folder
(c) New → folder
(d) File → New → folder
Answer:
(d) File → New → folder

Question 33.
Applications or files or folders are opened using related shortcut icons by ……………….
(a) Click and drag
(b) double click
(c) click
(d) drag and drop
Answer:
(b) double click

Question 34.
In windows 7, which option is used from file menu to quit an application?
(a) Exit
(b) Close
(c) Quit
(d) Exit window
Answer:
(b) Close

Question 35.
Which option is used to save the file?
(a) Ctrl + s
(b) Save
(c) File + save
(d) All the above
Answer:
(d) All the above

Question 36.
Which is inbuilt Word Processor application to creat and manipulate text documents?
(a) Word pad
(b) MS – word
(c) Staroffice writer
(d) Notepad
Answer:
(a) Word pad

Question 37.
ubuntu supports office suite called ……………….
(a) Open Office
(b) Star Office
(c) Libre Office
(d) MS – Office
Answer:
(c) Libre Office

Question 38.
The in – built e – mail software facility in ubuntu is ……………….
(a) Debain
(b) Thunderbird
(c) Gmail
(d) Firefox
Answer:
(b) Thunderbird

Question 39.
Which one of the following is a server distribution of Linux?
(a) Deepin
(b) Firefox
(c) MS.word
(d) Files
Answer:
(a) Deepin

Question 40.
The popular Linux distributions is ……………….
(a) Deepin
(b) Centos
(c) Debian
(d) All the above
Answer:
(d) All the above

Question 41.
Which option is used to delete all files in the Recycle bin?
(a) Remove the Recycle bin
(b) Empty the Recycle bin
(c) Clear the Recycle bin
(d) Clean the Recycle bin
Answer:
(b) Empty the Recycle bin

Question 42.
The search text box in the computer disk drive screen will appear at ……………….
(a) Bottom right comer
(b) Top left comer
(c) Bottom left comer
(d) Top right comer
Answer:
(d) Top right comer

Question 43.
Which key is used to access the menu’s in the menu bar?
(a) shift
(b) control
(c) alt
(d) Tab
Answer:
(c) alt

Question 44.
Which one of the following is the open source for desktop and server?
(a) Linux
(b) MS – DOS
(c) BASIC
(d) COBOL
Answer:
(a) Linux

Question 45.
Which mouse actions is used to display popup menu?
(a) right click
(b) click
(c) Double click
(d) drag and drop
Answer:
(a) right click

Question 46.
What is used to interact with windows by clicking icons?
(a) Mouse
(b) Keyboard
(c) Monitor
(d) Printer
Answer:
(a) Mouse

Question 47.
Which menu contains layout options?
(a) option
(b) view
(c) organize
(d) Menu bar
Answer:
(c) organize

Question 48.
V Box stands for ……………….
(a) Virtual Image Box
(b) Virtual box
(c) Validity box
(d) Vat Box
Answer:
(b) Virtual box

Question 49.
Session Indicator is present in which comer?
(a) top left
(b) top right
(c) bottom left
(d) bottom right
Answer:
(b) top right

Question 50.
Hardware settings is used in which option?
(a) Monitor
(b) Display
(c) Theme
(d) My Computer
Answer:
(b) Display

Question 51.
In Text entry settings En, Fr, Ku are ……………….
(a) Desktop Layouts
(b) Keyboard Layouts
(c) Message Layouts
(d) Data Entry Layouts
Answer:
(b) Keyboard Layouts

Question 52.
Which one in ubuntu is similar to task bar of windows?
(a) Launcher
(b) Desktop
(c) start
(d) Notification area
Answer:
(a) Launcher

Question 53.
Notification area of ubuntu desktop is otherwise called as ……………….
(a) Task bar
(b) Desktop
(c) status bar
(d) Indicator area
Answer:
(d) Indicator area

Question 54.
The indicator that helps to access instant messenger and email client is ……………….
(a) Messaging
(b) Network
(c) Email
(d) Text entry
Answer:
(a) Messaging

Question 55.
The application in Ubuntu similar to MS – Excel in windows ……………….
(a) Spread sheet
(b) Libre Office calc
(c) Open Office calc
(d) Star Office calc
Answer:
(b) Libre Office calc

Question 56.
Which dropdown list box is used to select the location where you want to save the file?
(a) Location
(b) Look – in
(c) Peep in
(d) Search for
Answer:
(b) Look – in

Question 57.
Which menu has the rename option?
(a) File
(b) Edit
(c) View
(d) Window
Answer:
(a) File

Question 58.
How will you rename the file?
(a) Edit → Rename
(b) press F2
(c) right click → rename
(d) All the above
Answer:
(d) All the above

Question 59.
In which panel of disk drive window, the files and folders are displayed in tree like structures?
(a) Top
(b) Centre
(c) Left
(d) Right
Answer:
(c) Left

Question 60.
Delete option is present in which menu?
(a) File
(b) Edit
(c) View
(d) Tools
Answer:
(a) File

Question 61.
Which option reboot the computer?
(a) Restart
(b) Boot
(c) Reboot
(d) Reselect
Answer:
(a) Restart

Question 62.
Identify the menu item which is not present is the keyboard indicator menu?
(a) Character Map
(b) Keyboard Layout
(c) Keyboard Layout Chart
(d) Text entry settings
Answer:
(b) Keyboard Layout

Question 63.
There are ………………. types of indicators in the Menu bar.
(a) 5
(b) 6
(c) 1
(d)8
Answer:
(b) 6

Question 64.
What is the name of the default Ubuntu 16.04 theme?
(a) Ambiance
(b) V Box
(c) Trash
(d) Files
Answer:
(a) Ambiance

Question 65.
Who developed Ubuntu OS?
(a) Mark Shuttleworth
(b) Ricki Mascitti
(c) Dan Bricklin
(d) Bob Franston
Answer:
(a) Mark Shuttleworth

Question 66.
Which year ubuntu OS was developed?
(a) 2003
(b) 2004
(c) 2005
(d) 2006
Answer:
(b) 2004

Question 67.
Identify the statement which is given wrong?
(a) Open source code is available free of cost.
(b) ubuntu is a linux based OS.
(c) ubuntu has smart searching facility
(d) The horizontal bar of icons on the left side of the desktop is called Launcher.
Answer:
(d) The horizontal bar of icons on the left side of the desktop is called Launcher.

Question 68.
How many ways of creating files are there is windows?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2

Question 69.
How many sets of scroll bars are there?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2

Question 70.
How many versions of windows 2000 are there?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(c) 4

Question 71.
Which one of the following is used for high traffic computer networks in windows 2000?
(a) Professional
(b) Server
(c) Advanced server
(d) Data centre server
Answer:
(d) Data centre server

Question 72.
How many types of Icons are there?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

Question 73.
What is the name given to the larger window?
(a) Work window
(b) Document window
(c) Application window
(d) Desktop
Answer:
(c) Application window

Question 74.
Which symbol is used to indicate that sub menu is attached to this option?
(a) +
(b) ▷
(c) ☐
(d) ◁
Answer:
(b) ▷

Question 75.
What is the keyboard shortcut for Exit option?
(a) Ctrl + E
(b) Alt + E
(c) Ctrl + Q
(d) Alt + Q
Answer:
(c) Ctrl + Q

Question 76.
How many methods of Renaming file are there?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

Question 77.
Which one of the following is not a method of pasting the contents?
(a) Edit → paste
(b) Ctrl + V
(c) alt + V
(d) Right click → paste
Answer:
(c) alt + V

Question 78.
Which option is used as a part of installing new software or windows update?
(a) Lock
(b) Restart
(c) Sleep
(d) Hibernate
Answer:
(b) Restart

Question 79.
Which option is found only on Laptop?
(a) Lock
(b) Restart
(c) Sleep
(d) Hibernate
Answer:
(d) Hibernate

PART – 2
II. Short Answers

Question 1.
Name some distributions of Linux.
Answer:

  1. Fedora
  2. Ubuntu
  3. BOSS
  4. RedHat
  5. Linux Mint.

Question 2.
What is Ubuntu?
Answer:
It is a Linux – based operating system. Designed for computers, smart phones and network servers.

Question 3.
What is Open source Operating system?
Answer:
It is the software in which the source code is available to the general public for use and modification from its original design, free of charge.

Question 4.
Mention any 2 significant features of Ubuntu?
Answer:

  1. It supports the office suite called Libreoffice.
  2. Ubuntu has in built email software called Thunderbird, which gives the user access to email such as Exchange, Gmail, Hotmail, etc.

Question 5.
What are the similarities between Ubuntu and other operating systems?
Answer:
All are based on the concepts of Graphical User Interface.

Question 6.
What is Multitasking?
Answer:
Multiple applications which can be executed simultaneously in Windows is known as Multitasking.

Question 7.
What is a workspace of a window?
Answer:
The workspace is the area in the document window to enter or type the text of your document. The workspace is the element of a window.

Question 8.
What is a Wordpad?
Answer:
It is an in – built word processor application in windows OS to create and manipulate text windows.

Question 9.
What is Folder?
Answer:
It is a container of files.

Question 10.
What is Firefox?
Answer:
It is a browser.

PART – 3
III. Explain in Brief

Question 1.
How will you renaming the files or folders?
Answer:
Using the FILE Menu

  1. Select the File or Folder you wish to Rename.
  2. Click File → Rename.
  3. Type in the new name.
  4. To finalise the renaming operation, press Enter

Question 2.
Draw the diagram of overview of an operating system
Answer:
Samacheer Kalvi 11th Computer Science Solutions Chapter 5 Working with Typical Operating System (Windows & Linux)

Question 3.
List out the important functions of an OS?
Answer:
Following are some of the important functions of an Operating System:

  1. Memory Management
  2. Process Management
  3. Device Management
  4. File Management
  5. Security Management
  6. Control overall system performance
  7. Error detecting aids
  8. Coordination between other software and users

Question 4.
How will you create a shortcuts on the desktop?
Answer:
Creating Shortcuts on the Desktop:
Shortcuts to your most often used folders and files may be created and placed on the Desktop to help automate your work.

  1. Select the file or folder that you wish to have as a shortcut on the Desktop.
  2. Right click on the file or folder.
  3. Select Send to from the shortcut menu, then select Desktop (create shortcut) from the submenu.
  4. A shortcut for the file or folder will now appear on your desktop and you can open it from the desktop in the same way as any other icon.

Question 5.
What are the elements of Ubuntu?
Answer:

  1. Search your Computer Icon
  2. Files
  3. Firefox Web browser
  4. Libre Office Writer
  5. Libre Office Calc
  6. Libre office Impress
  7. Ubuntu software icon
  8. Online shopping icon
  9. System settings icon
  10. Trash

Question 6.
Write note on windows XP.
Answer:

  1. Windows XP was developed in the year 2001
  2. It introduced 64 bit processor.
  3. Improved windows appearance with themes and offered a stable version.

Question 7.
What are the four versions of windows 2000.
Answer:

  1. Professional – business desktop and laptop systems.
  2. Server – used as both web server and an office server.
  3. Advanced server – for line of business applications.
  4. Data center server – for high-traffic computer networks.

Question 8.
What are the advantages of using windows 8 OS?
Answer:

  1. Windows 8 is faster than previous versions of windows.
  2. It has multi – core processing, Solid State Drivers (SSD), touch screens and other alternate input methods.
  3. It served as common platform for mobile and computer.

Question 9.
What is a Window?
Answer:
Window is a typical rectangular area in an application or a document. It is an area on the screen that displays information for a specific program. The two types of windows are application window and document window.

Question 10.
Differentiate Application Window and Document Window?
Answer:
Application Window:

  • It is the larger window.
  • This window helps the user to communicate with the application program.

Document Window:

  • It is the smaller window and appears inside the Application Windows.
  • This window is used for typing, editing, drawing and formatting the text and graphics.

Question 11.
How will you start an application?
Answer:

  1. Click the start button and then point to all programs. The program menu appears. Point to the group that contains the application you want to start and then click the application name.
  2. Click Run on the start menu and type the name of the application and click ok.

Question 12.
How will you close an application?
Answer:

  1. To quit a application, click the close button in the upper right comer of the application window.
  2. Click File → Exit
  3. Click File → Close option

Question 13.
How will you open the word pad?
Answer:

  1. Click start → All programs → Accessories → Wordpad
  2. Run → type wordpad, click ok

Question 14.
How will you delete a file or folder?
Answer:
Select the file or folder you wish to delete.

  1. Right click the file or folder, select delete option from the popup menu or click file → delete (or) Press delete key from the keyboard.
  2. The file will be deleted and moved to the Recycle bin.

Question 15.
Write note on Recycle bin.
Answer:
Recycle bin is a special folder to keep the files or folders deleted by the user. From the recycle bin, files can be restored back.

PART – 4
IV. Explain in Detail

Question 1.
Discuss about the mouse actions.
Answer:

Action Reaction
Point to an item Move the mouse pointer over the item.
Click Point to the item on the screen, press and release the left mouse button.
Right click Point to the item on the screen, press and release the right mouse button. Clicking the right mouse button displays a pop up menu with various options.
Double-click Point to the item on the screen, quickly press twice the left mouse button.
Drag and drop Point to an item then hold the left mouse button as you move the pointer press and you have reached the desired position, release the mouse button.

Question 2.
Discuss the following Icons.
Answer:
Shortcut Icons:
Shortcut icons can be created for any application or file or folder. By double clicking the icon, the related application or file or folder will open. This represents the shortcut to open a particular application.

Disk drive icons:
The disk drive icons graphically represent five disk drive options.

  1. Hard disk.
  2. CD – ROM/DVD Drive.
  3. Pen drive.
  4. Other removable storage such as mobile, smart phone, tablet etc.
  5. Network drives if your system is connected with other system.

Question 3.
How will you copying files and folders to removable disk?
Answer:
Copying Files and Folders to removable disk:
There are several methods of transferring files to or from a removable disk.

  1. Copy and Paste
  2. Send To

METHOD I – Copy and Paste:

  1. Plug the USB flash drive directly into an available USB port.
  2. If the USB flash drive or external drive folder does NOT open automatically, follow these steps:
  3. Click Start → Computer.
    Samacheer Kalvi 11th Computer Science Solutions Chapter 5 Working with Typical Operating System (Windows & Linux)
  4. Double Clicking on the Removable Disk associated with the USB flash drive.
    Samacheer Kalvi 11th Computer Science Solutions Chapter 5 Working with Typical Operating System (Windows & Linux)
  5. Navigate to the folders in your computer containing files you want to transfer. Right – click on the file you want to copy, then select Copy.
    Samacheer Kalvi 11th Computer Science Solutions Chapter 5 Working with Typical Operating System (Windows & Linux)
  6. Return to the Removable Disk window, right – click within the window, then select Paste.
    Samacheer Kalvi 11th Computer Science Solutions Chapter 5 Working with Typical Operating System (Windows & Linux)

METHOD II – Send To:

  1. Plug the USB flash drive directly into an available USB port.
  2. Navigate to the folders in your computer containing files you want to transfer.
  3. Right – click on the file you want to transfer to your removable disk.
  4. Click Send To and select the Removable Disk associated with the USB flash drive.
    Samacheer Kalvi 11th Computer Science Solutions Chapter 5 Working with Typical Operating System (Windows & Linux)

Question 4.
Explain the various elements of a window.
Answer:
Elements of a window:
Samacheer Kalvi 11th Computer Science Solutions Chapter 5 Working with Typical Operating System (Windows & Linux)
Title Bar : The title bar will display the name of the application and the name of the document opened. It will also contain minimize, maximize and close button.

Menu Bar : The menu bar is seen under the title bar. Menus in the menu bar can be accessed by pressing Alt key and the letter that appears underlined in the menu title. Additionally, pressing Alt or F10 brings the focus on the first menu of the menu bar. In Windows 7, in the absence of the menu bar, click Organise and from the drop down menu, click the Layout option and select the desired item, from that list.

The Workspace : The workspace is the area in the document window to enter or type the text of your document.

Scroll bars : The scroll bars are used to scroll the workspace horizontally or vertically.

Corners and borders : The comers and borders of the window helps to drag and resize the windows. The mouse pointer changes to a double headed arrow when positioned over a border or a comer. Drag the border or comer in the direction indicated by the double headed arrow to the desired size as shown in the Figure. The window can be resized by dragging the comers diagonally across the screen.

Question 5.
Explain various methods of creating Files and Folders.
Answer:
Creating files and Folders:
Creating Folders:
You can store your files in many locations – on the hard disk or in other devices. To better organise your files, you can store them in folders.
There are two ways in which you can create a new folder:

Method I:
Step 1: Open Computer Icon.
Step 2: Open any drive where you want to create a new folder. (For example select D:)
Step 3: Click on File → New → Folder.
Step 4: A new folder is created with the default name “New folder”.
Step 5: Type in the folder name and press Enter key.

Method II:
In order to create a folder in the desktop:
Step 1: In the Desktop, right click → New → Folder.
Step 2: A Folder appears with the default name “New folder” and it will be highlighted.
Step 3: Type the name you want and press Enter Key.
Step 4: The name of the folder will change.

Question 6.
Explain how will you rename files and folders.
Answer:
Renaming Files or Folders:
There are number of ways to rename files or folders. You can rename using the File menu, left mouse button or right mouse button.

Method I:
Using the FILE Menu

  1. Select the File or Folder you wish to Rename.
  2. Click File → Rename.
  3. Type in the new name.
  4. To finalise the renaming operation, press Enter.

Method II:
Using the Right Mouse Button

  1. Select the file or folder you wish to rename.
  2. Click the right mouse button over the file or folder.
  3. Select Rename from the pop-up menu. ’
  4. Type in the new name.
  5. To finalise the renaming operation, press Enter.
  6. The folder “New Folder” is renamed as C++.

Method III:
Using the Left Mouse Button

  1. Select the file or folder you wish to rename.
  2. Press F2 or click over the file or folder. A surrounding rectangle will appear around the name.
  3. Type in the new name.
  4. To finalise the renaming operation, press Enter.

Question 7.
How will you move the file or folder from one place to another?
Answer:
Moving/Copying Files and Folders:
You can move your files or folders to other areas using variety of methods.

Moving Files and Folders:
Method I – CUT and PASTE
To move a file or folder, first select the file or folder and then choose one of the following:

  1. Click on the Edit → Cut or Ctrl + X Or right click → cut from the pop – up menu.
  2. To move the file(s) or folder(s) in the new location, navigate to the new location and paste it using Click Edit → Paste from edit menu or Ctrl + V using keyboard.
  3. Or Right click → Paste from the popup menu. The file will be pasted in the new location.

Method II – Drag and Drop:

In the disk drive window, we have two panes called left and right panes. In the left pane, the files or folders are displayed like a tree structure. In the right pane, the files inside the specific folders in the left pane are displayed with various options.

  1. In the right pane of the Disk drive window, select the file or folder you want to move.
  2. Click and drag the selected file or folder from the right pane, to the folder list on the left pane.
  3. Release the mouse button when the target folder is highlighted (active).
  4. Your file or folder will now appear in the new area.

Question 8.
How will you copy files and folders.
Answer:
Copying Files and Folders:
There are variety of ways to copy files and folders:

Method I – COPY and PASTE:
To copy a file or folder, first select the file or folder and then choose one of the following:

  1. Click Edit → Copy or Ctrl + C or right click → Copy from the pop – up menu.
  2. To paste the file(s) or folder(s) in the new location, navigate to the target location then do one of the following:
  3. Click Edit → Paste or Ctrl + V.
  4. Or Right click → Paste from the pop – up menu.

Method II – Drag and Drop:

  1. In the RIGHT pane, select the file or folder you want to copy.
  2. Click and drag the selected file and/or folder to the folder list on the left, and drop it where you want to copy the file and/or folder.
  3. Your file(s) and folder(s) will now appear in the new area.

Question 10.
Explain various menu bar Indicator.
Menu bar:
The menu bar is located at the top of the screen. The menu bar incorporates common functions used in Ubuntu. The frequently used icons in the menu bar are found on the right. The most common indicators in the Menu bar are located in the indicator or notification area.

Network indicator:
This manages network connections, allowing you to connect to a wired or wireless network. Text entry settings:
This shows the current keyboard layout (such as Eli, Fr, Ku, and so on). If more than one keyboard layout is shown, it allows you to select a keyboard layout out of those choices. The keyboard indicator menu contains the following menu items: Character Map, Keyboard Layout Chart, and Text Entry Settings.

Messaging indicator:
This incorporates your social applications. From here, you can access instant messenger and email clients.

Sound indicator:
This provides an easy way to adjust the volume as well as access your music player.

Clock:
This displays the current time and provides a link to your calendar and time and date settings.

Question 11.
Explain various elements of Ubuntu.
Answer:
Samacheer Kalvi 11th Computer Science Solutions Chapter 5 Working with Typical Operating System (Windows & Linux) 1
Search your Computer Icon : This icon is equal to search button in Windows OS. Here, you have to give the name of the File or Folder for searching them.

Files : This icon is equivalent to My Computer icon. From here, you can directly go to Desktop, Documents and so on.

Firefox Web Browser : By clicking this icon, you can directly browse the internet. This is equivalent to clicking the Web Browser in Task bar in Windows.

Libre Office Writer : This icon will directly take you to document preparation application like MS Word in Windows.

Libre Office Calc : This icon will open LibreOffice Calc application. It is similar to MS Excel in Windows.

Libre Office Impress : By clicking this icon, you can open LibreOffice Impress to prepare any presentations in Ubuntu like MS PowerPoint.

Ubuntu Software Icon : This icon will let you add any additional applications you want. This can be done by clicking the Update option at the top right comer of that screen.

Online Shopping icon : Using this icon user can buy and sell any products online.

System Settings Icons : This icon is similar to the Control panel in the Windows Operating System. But here, you need to authenticate the changes by giving your password. You cannot simply change as you do in Windows.

Trash : This icon is the equivalent of Recycle bin of windows OS. All the deleted Files and Folders are moved here.

Samacheer Kalvi 11th Accountancy Solutions Chapter 7 Subsidiary Books – II

Students can Download Accountancy Chapter 7 Subsidiary Books – II Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Accountancy Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Accountancy Solutions Chapter 7 Subsidiary Books – II

Samacheer Kalvi 11th Accountancy Subsidiary Books – II Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the Correct Answer

11th Accountancy Chapter 7 Book Back Answers Question 1.
Cash book is a ………………
(a) Subsidiary book
(b) Principal book
(c) Journal proper
(d) Both subsidiary book and principal book
Answer:
(d) Both subsidiary book and principal book

Subsidiary Books Questions And Answers Pdf Question 2.
The cash book records ………………
(a) All cash receipts
(b) All cash payments
(c) Both (a) and (b)
(d) All credit transactions
Answer:
(c) Both (a) and (b)

11th Accountancy 7th Chapter Solutions Question 3.
When a firm maintains a simple cash book, it need not maintain ………………
(a) Sales account in the ledger
(b) Purchases account in the ledger
(c) Capital account in the ledger
(d) Cash account in the ledger
Answer:
(d) Cash account in the ledger

11th Accountancy Book Back Answers Question 4.
A cash book with discount, cash and bank column is called ………………
(a) Simple cash book
(b) Double column cash book
(c) Three column cash book
(d) Petty cash book
Answer:
(c) Three column cash book

11th Accountancy – Book Back Answers Question 5.
In Triple column cash book, the balance of bank overdraft brought forward will appear in ………………
(a) Cash column debit side
(b) Cash column credit side
(c) Bank column debit side
(d) Bank column credit side
Answer:
(d) Bank column credit side

Samacheer Kalvi 11th Accountancy Chapter 7 Question 6.
Which of the following is recorded as contra entry?
(a) Withdrew cash from bank for personal use
(b) Withdrew cash from bank for office use
(c) Direct payment by the customer in the bank account of the business
(d) When bank charges interest
Answer:
(b) Withdrew cash from bank for office use

Samacheer Kalvi Guru 11th Accountancy Question 7.
If the debit and credit aspects of a transaction are recorded in the cash book, it is ………………
(a) Contra entry
(b) Compound entry
(c) Single entry
(d) Simple entry
Answer:
(a) Contra entry

Accounting 11 Chapter 7 Answer Key Question 8.
The balance in the petty cash book is ………………
(a) An expense
(b) A profit
(c) An asset
(d) A liability
Answer:
(c) An asset

Class 11 Accountancy Chapter 7 Solutions Question 9.
Petty cash may be used to pay ………………
(a) The expenses relating to postage and conveyance
(b) Salary to the Manager
(c) Purchase of furniture and fixtures
(d) Purchase of raw materials
Answer:
(a) The expenses relating to postage and conveyance

Accounting Chapter 7 Answer Key Question 10.
Small payments are recorded in a book called ………………
(a) Cash book
(b) Purchase book
(c) Bills payable book
(d) Petty cash book
Answer:
(d) Petty cash book

II. Very Short Answer Questions

11th Accountancy Book Answers Question 1.
What is cash book?
Answer:
Cash book is the book in which only cash transactions are recorded in the chronological order. The cash book is the book of original entry or prime entry as cash transactions are recorded for the first time in it. Cash transactions here may include bank transactions also. Cash receipts are recorded on the debit side while cash payments are recorded on the credit side.

Class 11 Accounts Subsidiary Books Solutions Question 2.
What are the different types of cash book?
Answer:
The main cash book may be of various types and following are the three most common types.

  1. Simple or single column cash book (only cash column)
  2. Cash book with cash and discount column (double column cash book)
  3. Cash book with cash, discount and bank columns (three column cash book)

Accountancy Class 11 Chapter 7 Solutions Question 3.
What is simple cash book?
Answer:
Single column cash book or simple cash book, like a ledger account has only one amount column, i.e., cash column on each side. Only cash transactions are recorded in this book. All cash receipts and payments are recorded systematically in this book.

Chapter 7 Accountancy Class 11 Question 4.
Give the format of ‘Single column cash book’.
Answer:
Simple Cash Book
Subsidiary Books Questions And Answers Pdf Samacheer Kalvi 11th Accountancy Solutions Chapter 7

Subsidiary Books Class 11 Notes Question 5.
What is double column cash book?
Answer:
It is a cash book with cash and discount columns. As there are two columns, i.e., discount and cash columns, both on debit and credit sides, this cash book is known as ‘double column cash book’.

Question 6.
Give the format of ‘Double column cash book’.
Answer:
Cash book with cash and discount columns
11th Accountancy 7th Chapter Solutions Samacheer Kalvi Subsidiary Books

Question 7.
What is three column cash book?
Answer:
A three column cash book includes three amount columns on both sides, i.e., cash, bank and discount. This cash book is prepared in the same way as simple and double column cash books are prepared. The transactions which increase the cash and bank balance are recorded on the debit side of the cash and bank columns respectively. Opening balance of cash and favourable bank balance appear as the first item on the debit side of the three column cash book in case of existing business. If the business is a new one, capital contributed in cash and/or bank deposit appear as the first item on the debit side.

Question 8.
What is cash discount?
Answer:
Cash discount is allowed to the parties making prompt payment within the stipulated period of time or early payment. It is discount allowed (loss) for the creditor and discount received (gain) for the debtor who makes payment. The discount is allowed when payment is received or made and hence, the entry for discount is also passed with the entry of payment. The earlier the payment, the more may be the discount. Cash discount motivates the debtor to make the payment at an earlier date to avail discount facility.

Question 9.
What is trade discount?
Answer:
Trade discount is a deduction given by the supplier to the buyer on the list price or catalogue price of the goods. It is given as a trade practice or when goods are purchased in large quantities. It is shown as a deduction in the invoice. Trade discount is not recorded in the books of accounts. Only the net amount is recorded.

Question 10.
What is a petty cash book?
Answer:
Business entities have to pay various small expenses like taxi fare, bus fare, postage, carriage, stationery, refreshment and other sundry items. These are small payments and repetitive in nature. If all these small payments are recorded in the main cash book, it will be loaded with lot of entries. Hence, all petty payments of the business may be recorded in a separate book, which is called as petty cash book and the person who maintains the petty cash book is called the petty cashier.

III. Short Answer Questions

Question 1.
Explain the meaning of imprest system of petty cash book.
Answer:
Under this system, a fixed amount necessary or sufficient to meet petty payments determined on the basis of past experience is paid to the petty cashier on the first day of the period. (It may be a week or fortnight or month). The amount given to the petty cashier in advance is known as “Imprest Money”.

The word imprest means payment in advance. The petty cashier makes payments from this amount and records them in petty cash book. At the end of a particular period, the petty cashier submits the petty cash book to the head cashier.

The head cashier scrutinises the petty payments and gives amount equal to the amount spent by petty cashier so that the total amount with the petty cashier is now equal to the amount he had received in the beginning as advance. Under the system, the total cash with the petty cashier never exceeds the imprest at any time during the period.

Question 2.
Bring out the differences between cash discount and trade discount.
Answer:
Following are the difference between cash discount and trade discount:
11th Accountancy Book Back Answers Samacheer Kalvi Chapter 7 Subsidiary Books

Question 3.
Write the advantages of maintaining petty cash book.
Answer:
Following are the advantages of maintaining petty cash book:

  1. There can be better control over petty payments.
  2. There is saving of time of the main cashier.
  3. Cash book is not loaded with many petty payments.
  4. Posting of entries from main cash book and petty cash book is comparatively easy.

Question 4.
Write a brief note on accounting treatment of discount in cash book.
Answer:
In the discount columns, cash discount, i.e., cash discount allowed and cash discount received are recorded. The net amount received is entered in the amount column on the debit side and the net amount paid is entered in the amount column on the credit side. For the seller who allows cash discount, it is a loss and hence it is debited and shown on the debit side of the cash book. For the person making payment, discount received is a gain because less payment is made and it is credited and shown on the credit side of the cash book.

Question 5.
Briefly explain about contra entry with examples.
Answer:
To denote that there are contra entries, the alphabet ‘C’ is written in L.F. column on both sides. Contra means that particular entry is posted on the other side (contra) of the same book, because Cash account and Bank account are there in the cash book only and there are no separate ledger accounts needed for this purpose. The alphabet ‘C’ indicates that no further posting is required and the relevant account is posted on the opposite side.

IV. Exercises

Question 1.
Enter the following transactions in a single column cash book of Seshadri for May, 2017.
11th Accountancy - Book Back Answers Samacheer Kalvi Chapter 7 Subsidiary Books
Answer:
In the books of Seshadri
Single column cash book
Samacheer Kalvi 11th Accountancy Chapter 7 Subsidiary Books

Question 2.
Enter the following transactions in a single column cash book of Pandeeswari for the month ofJune, 2017
Samacheer Kalvi Guru 11th Accountancy Solutions Chapter 7 Subsidiary Books
Answer:
In the books of Pandeeswari
Single column cash book
Accounting 11 Chapter 7 Answer Key Samacheer Kalvi Subsidiary Books – II

Question 3.
Enter the following transactions in a single column cash book of Ramalingam for month of July, 2017.
Class 11 Accountancy Chapter 7 Solutions Samacheer Kalvi Subsidiary Books – II
Answer:
In the books of Ramalingam
Single column cash book
Accounting Chapter 7 Answer Key Samacheer Kalvi Subsidiary Books – II

Question 4.
Enter the following transaction in Ahamad’s cash book with discount and cash columns.
11th Accountancy Book Answers Samacheer Kalvi Chapter 7 Subsidiary Books – II
Answer:
In the books of Ahamad’s
Cash book with discount and cash columns
Class 11 Accounts Subsidiary Books Solutions Samacheer Kalvi Chapter 7

Question 5.
Enter the following transaction in Chandran’s cash book with cash and discount column.
Accountancy Class 11 Chapter 7 Solutions Samacheer Kalvi Subsidiary Books – II
Answer:
In the books of chandran
Cash book with cash and discount columns
Chapter 7 Accountancy Class 11 Samacheer Kalvi Subsidiary Books – II

Question 6.
Enter the following transactions in cash book with discount and cash column of Anand
Subsidiary Books Class 11 Notes Samacheer Kalvi Accountancy Solutions Chapter 7
Answer:
In the books of Anand
Cash book with cash and discount and cash columns
Samacheer Kalvi 11th Accountancy Solutions Chapter 7 Subsidiary Books – II

Question 7.
Write out a cash book with discount, cash and bank columns in the books of Mahendran.
Samacheer Kalvi 11th Accountancy Solutions Chapter 7 Subsidiary Books – II
Answer:
In the books of Mahendran
There – column cash book
Samacheer Kalvi 11th Accountancy Solutions Chapter 7 Subsidiary Books – II

Question 8.
Enter the following transactions in the three column cash book of Kalyana Sundaram
Samacheer Kalvi 11th Accountancy Solutions Chapter 7 Subsidiary Books – II
Answer:
In the books of Kalyana Sundaram
There – column cash book
Samacheer Kalvi 11th Accountancy Solutions Chapter 7 Subsidiary Books – II

Question 9.
Enter the following transactions of Fathima in the cash book with cash, bank and discount columns for the month of May, 2017.
Samacheer Kalvi 11th Accountancy Solutions Chapter 7 Subsidiary Books – II
Answer:
In the books of Fathima
Three – column cash book with cash, bank and discount
Samacheer Kalvi 11th Accountancy Solutions Chapter 7 Subsidiary Books – II
Question 10.
Enter the following transactions in the three column cash book of Chozhan.
Samacheer Kalvi 11th Accountancy Solutions Chapter 7 Subsidiary Books – II
Answer:
In the books of Fathima
Three – column cash book with cash, bank and discount
Samacheer Kalvi 11th Accountancy Solutions Chapter 7 Subsidiary Books – II

Question 11.
Enter the following transactions in a cash book with cash, bank and discount columns of Sundari.
Samacheer Kalvi 11th Accountancy Solutions Chapter 7 Subsidiary Books – II
Answer:
In the books of Sundari
Three – column cash book
Samacheer Kalvi 11th Accountancy Solutions Chapter 7 Subsidiary Books – II

Question 12.
Record the following transactions in the three column cash book of Rajeswari for the month of June, 2017.
Samacheer Kalvi 11th Accountancy Solutions Chapter 7 Subsidiary Books – II
Answer:
In the books of Rajeswari
Three – column cash book
Samacheer Kalvi 11th Accountancy Solutions Chapter 7 Subsidiary Books – II

Question 13.
Record the following transactions in three column cash book of Ramachandran.
Samacheer Kalvi 11th Accountancy Solutions Chapter 7 Subsidiary Books – II
Answer:
In the books of Ramachandran
Three – column cash book
Samacheer Kalvi 11th Accountancy Solutions Chapter 7 Subsidiary Books – II

Question 14.
Record the following transactions in the three column cash book of John Pandian.
In the books of John Pandian
Three – column cash book
Samacheer Kalvi 11th Accountancy Solutions Chapter 7 Subsidiary Books – II
Answer:
In the books of John Pandian
Three – column cash book
Samacheer Kalvi 11th Accountancy Solutions Chapter 7 Subsidiary Books – II
Question 15.
Prepare a triple column cash book of Rahim from the following transactions:
Samacheer Kalvi 11th Accountancy Solutions Chapter 7 Subsidiary Books – II
In the hooks of Rahim
Three – column cash book
Answer:
Samacheer Kalvi 11th Accountancy Solutions Chapter 7 Subsidiary Books – II

Question 16.
Prepare analytical petty cash book from the following particulars under imprest system:
Samacheer Kalvi 11th Accountancy Solutions Chapter 7 Subsidiary Books – II
Answer:
Analytical Petty Cash Book (in ₹)
Samacheer Kalvi 11th Accountancy Solutions Chapter 7 Subsidiary Books – II

Question 17.
From the following information prepare an analytical petty cash book under imprest system:
Samacheer Kalvi 11th Accountancy Solutions Chapter 7 Subsidiary Books – II
Answer:
Analytical Petty Cash Book (in ₹)
Samacheer Kalvi 11th Accountancy Solutions Chapter 7 Subsidiary Books – II

Question 18.
Record the following transactions in an analytical petty cash book and balance the same. On 1st November, 2017, the petty cashier started with imprest cash ₹ 2,000.
Samacheer Kalvi 11th Accountancy Solutions Chapter 7 Subsidiary Books – II
Answer:
Analytical Petty Cash Book (in ₹)
Samacheer Kalvi 11th Accountancy Solutions Chapter 7 Subsidiary Books – II

Question 19.
Enter the following transactions in Iyyappan’s petty cash book with analytical columns under imprest system.
Samacheer Kalvi 11th Accountancy Solutions Chapter 7 Subsidiary Books – II
Answer:
Analytical Petty Cash Book (in ₹)
Samacheer Kalvi 11th Accountancy Solutions Chapter 7 Subsidiary Books – II

Textbook Case Study Solved

Vetri is a sole trader selling food products. He maintains a simple cash book. He sells and purchases goods both on cash and credit. He maintains the cash book by himself. He allows discount and receives discount. He has his personal bank account. He also has so many petty expenses. Now, he wants to establish his business. But he wants to maintain the cash book all by himself.
Now, discuss on the following points:

Question 1.
What could be the reason that Vetri maintains the cash book by himself?
Answer:
He is a sole trader, he need not show the accounts to anybody else, he wants to know about profit or loss for himself only. So he maintains cash book only.

Question 2.
Is it convenient for him to record all the cash transactions in the simple cash book?
Answer:
No, it is not convenient for him to record all the cash transactions in the simple cash book.

Question 3.
Will his personal bank account serve the purpose of his business transactions?
Answer:
No, his personal bank account will not serve the purpose of his business transactions.

Question 4.
Suggest him some better ways of maintaining the cash transactions.
Answer:
Suggestions:

He may maintain triple column cash book, because he can know all cash transactions is a same account.
Instead of personal bank account he can open business bank account (i.e.) current account.

Question 5.
When his business becomes large, what other books will he be maintaining?
Answer:
He will be maintaining the following other books:

  1. Triple column cash book.
  2. Petty cash book (Analytical).
  3. Purchases book for credit purchases
  4. Sales book for credit sales.
  5. Purchases returns book.
  6. Sales returns book.
  7. Business book account (i.e. current account to be maintained).
  8. Proper journal for other assets maintaining:
    • All cash transactions recorded in cash book.
    • All petty expenses are to recorded in analytical petty cash book.
    • All credit transactions to recorded in special purpose books (i.e. purchases book, sales book, purchases return book and sales returns book and proper journal.

Samacheer Kalvi 11th Accountancy Subsidiary Books – II Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer

Question 1.
Cash receipts are recorded on the ……………. of the cash book.
(a) debit side
(b) credit side
(c) journal
(d) ledger
Answer:
(a) debit side

Question 2.
Cash payments are recorded on the ……………. of the cash book.
(a) debit side
(b) credit side
(c) contra
(d) journal
Answer:
(b) credit side

Question 3.
All cash transactions are straightway recorded in the …………….
(a) cash book
(b) ledger
(c) journal
(d) contra
Answer:
(a) cash book

Question 4.
……………. serves the purpose of a journal and a ledger.
(a) cash book
(b) purchase book
(c) sales book
(d) petty cash book
Answer:
(a) cash book

Question 5.
There are ……………. types of cash book (common).
(a) three
(b) four
(c) five
(d) two
Answer:
(a) three

Question 6.
R.N. expands for …………….
(a) Receipts number
(b) Roll number
(c) Route number
(d) Rank number
Answer:
(a) Receipts number

Question 7.
L.F. expands for ……………..
(a) Long file
(b) Ledger folio
(c) Line folio
(d) Link file
Answer:
(b) Ledger folio

Question 8.
V.N. exapnds for …………….
(a) Value number
(b) Voucher number
(c) Vendor number
(d) Volume number
Answer:
(b) Voucher number

Question 9.
As both the debit and credit aspects of a transaction are recorded in the cash book, such entries are called ……………. entries.
(a) Contra
(b) Journal
(c) Compound
(d) Opening
Answer:
(a) Contra

Question 10.
The word ……………. means payment in advance.
(a) imprest
(b) loan
(c) money
(d) petty cash
Answer:
(a) imprest

Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement

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Samacheer Kalvi 11th Physics Nature of Physical World and Measurement TextBook Questions Solved

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Multiple Choice Questions

Samacheer Kalvi 11th Physics Solution Chapter 1 Question 1.
One of the combinations from the fundamental physical constants is \(\frac{h c}{\mathrm{G}}\). The unit of this expression is
(a) Kg2
(b) m3
(c) S-1
(d) m
Answer:
(a) Kg2

Nature Of Physical World And Measurement Question 2.
If the error in the measurement of radius is 2%, then the error in the determination of volume of the sphere will be …….
(a) 8%
(b) 2%
(c) 4%
(d) 6%
Answer:
(d) 6%

Nature Of Physical World And Measurement Questions And Answers Question 3.
If the length and time period of an oscillating pendulum have errors of 1 % and 3% respectively
then the error in measurement of acceleration due to gravity is …… [Related to AMPMT 2008]
(a) 4%
(b) 5%
(c) 6%
(d) 7%
Answer:
(d) 7%

11th Physics Nature Of Physical World And Measurement Question 4.
The length of a body is measured as 3.51 m, if the accuracy is 0.01mm, then the percentage error in the measurement is ……
(a) 351%
(b) 1%
(c) 0.28%
(d) 0.035%
Answer:
(c) 0.28%

Nature Of Physical World And Measurement Class 11 Question 5.
Which of the following has the highest number of significant figures?
(a) 0.007 m2
(b) 2.64 × 1024 kg
(c) 0.0006032 m2
(d) 6.3200 J
Answer:
(d) 6.3200 J

Nature Of Physical World And Measurement Class 11 Notes Question 6.
If π = 3.14, then the value of π2 is …..
(a) 9.8596
(b) 9.860
(c) 9.86
(d) 9.9
Answer:
(c) 9.86

Samacheer Kalvi Guru 11th Physics Question 7.
Which of the following pairs of physical quantities have same dimension?
(a) force and power
(b) torque and energy
(c) torque and power
(d) force and torque
Answer:
(b) torque and energy

11th Physics Unit 1 Question 8.
The dimensional formula of Planck’s constant h is [AMU, Main, JEE, NEET]
(a) [ML2T-1]
(b) [ML2T-3]
(c) [MLTT-1]
(d) [MLTT3-3]
Answer:
(a) [ML2T-1]

Physics Class 11 Chapter 1 Question 9.
The velocity of a particle v at an instant t is given by v = at + bt2. The dimensions of b is ……
(a) [L]
(b) [LT-1]
(c) [LT-2]
(d) [LT-3]
Answer:
(d) [LT-3]

Samacheerkalvi.Guru 11th Physics Question 10.
The dimensional formula for gravitational constant G is [Related to AIPMT 2004]
(a) [ML-3T-2]
(b) [M-1L3T-2]
(c) [M-1L-3T-2]
(d) [ML-3T2]
Answer:
(b) [M-1L3T-2]

Nature Of Physical World And Measurement In Tamil Question 11.
The density of a material in CGS system of units is 4 g cm-3. In a system of units in which unit of length is 10 cm and unit of mass is 100 g, then the value of density of material will be ……
(a) 0.04
(b) 0.4
(c) 40
(d) 400
Answer:
(c) 40

Samacheer Kalvi 11th Physics Question 12.
If the force is proportional to square of velocity, then the dimension of proportionality constant
is [JEE-2000] ……
(a) [MLT0]
(b) [MLT-1]
(c) [ML-2T]
(d) [ML-1T0]
Answer:
(d) [ML-1T0]

Samacheer Kalvi 11th Physics Solution Book Question 13.
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 1213 [MainAIPMT2011]
(a) length
(b) time
(c) velocity
(d) force
Answer:
(c) velocity

11 Th Physics Samacheer Kalvi Guide Question 14.
Planck’s constant (h), speed of light in vaccum (c) and Newton’s gravitational constant (G) are taken as three fundamental constants. Which of the following combinations of these has the dimension of length? [NEET 2016 (Phase II)]
Samacheer Kalvi 11th Physics Solution Chapter 1 Nature Of Physical World And Measurement
Answer:
Nature Of Physical World And Measurement Samacheer Kalvi 11th Physics Solutions Chapter 1

Samacheer Kalvi 11th Physics Guide Question 15.
A length-scale (l) depends on the permittivity (ε) of a dielectric material, Boltzmann constant (kB), the absolute temperature (T), the number per unit volume (n) of certain charged particles, and the charge (q) carried by each of the particles. Which of the following expression for l is dimensionally correct?. [JEE (advanced) 2016]
Nature Of Physical World And Measurement Questions And Answers Samacheer Kalvi 11th Physics Solutions Chapter 1
Answer:
11th Physics Nature Of Physical World And Measurement Samacheer Kalvi Chapter 1

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Short Answer Questions

11th Physics Chapter 1 Numerical Problems Question 1.
Briefly explain the types of physical quantities.
Answer:
Physical quantities are classified into two types. There are fundamental and derived quantities. Fundamental or base quantities are quantities which cannot be expressed in terms of any other physical quantities. These are length, mass, time, electric current, temperature, luminous intensity and amount of substance.
Quantities that can be expressed in terms of fundamental quantities are called derived quantities. For example, area, volume, velocity, acceleration, force.

11th Physics Samacheer Kalvi Question 2.
How will you measure the diameter of the Moon using parallax method?
Answer:
Let θ is the angular diameter of moon
d – is the distance of moon from earth, from figure, θ = \(\frac{\mathrm{D}}{d}\)
Nature Of Physical World And Measurement Class 11 Samacheer Kalvi Physics Solutions Chapter 1
Diameter of moon D = d.θ
by knowing θ, d, diameter of moon can be calculated

Samacheer Kalvi Physics 11th Question 3.
Write the rules for determining significant figures.
Answer:
Rules for counting significant figures:
Nature Of Physical World And Measurement Class 11 Notes Samacheer Kalvi Physics Solutions Chapter 1
Note: 1 Multiplying or dividing factors, which are neither rounded numbers nor numbers representing measured values, are exact and they have infinite numbers of significant figures as per the situation.
For example, circumference of circle S = 2πr, Here the factor 2 is exact number. It can be written as 2.0, 2.00 or 2.000 as required.
Note: 2 The power of 10 is irrelevant to the determination of significant figures.
For example x = 5.70 m = 5.70 × 102 cm = 5.70 × 103 mm = 5.70 × 10-3 km.
In each case the number of significant figures is three.

Physics Class 11 Samacheer Kalvi Question 4.
What are the limitations of dimensional analysis?
Answer:
Limitations of Dimensional analysis
1. This method gives no information about the dimensionless constants in the formula like 1, 2, ……… π, e, etc.
This method cannot decide whether the given quantity is a vector or a scalar.
This method is not suitable to derive relations involving trigonometric, exponential and logarithmic functions.
It cannot be applied to an equation involving more than three physical quantities.
It can only check on whether a physical relation is dimensionally correct but not the correctness of the relation. For example, using dimensional analysis, Samacheer Kalvi Guru 11th Physics Solutions Chapter 1 Nature Of Physical World And Measurement is dimensionally correct whereas the correct relation is 11th Physics Unit 1 Samacheer Kalvi Nature Of Physical World And Measurement

Samacheer Kalvi Guru 11 Physics Question 5.
Define precision and accuracy. Explain with one example.
Answer:
The accuracy of a measurement is a measure of how close the measured value is to the true value of the quantity. Precision of a measurement is a closeness of two or more measured values to each other.
The true value of a certain length is near 5.678 cm. In one experiment, using a measuring instrument of resolution 0.1 cm, the measured value is found to be 5.5 cm. In another experiment using a measuring instrument of greater resolution, say 0.01 cm, the length is found to be 5.38 cm. We find that the first measurement is more accurate as it is closer to the true value, but it has lesser precision. On the contrary, the second measurement is less accurate, but it is more precise.

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Long Answer Questions

Question 1.
(i) Explain the use of screw gauge and vernier caliper in measuring smaller distances.
Answer:
(i) Measurement of small distances: screw gauge and vernier caliper Screw gauge:
The screw gauge is an instrument used for measuring accurately the dimensions of objects up to a maximum of about 50 mm. The principle of the instrument is the magnification of linear motion using the circular motion of a screw. The least count of the screw gauge is 0.01 mm. Vernier caliper: A vernier caliper is a versatile instrument for measuring the dimensions of an object namely diameter of a hole, or a depth of a hole. The least count of the vernier caliper is 0.1 mm.
Physics Class 11 Chapter 1 Samacheer Kalvi Nature Of Physical World And Measurement
Samacheerkalvi.Guru 11th Physics Solutions Chapter 1 Nature Of Physical World And Measurement
Nature Of Physical World And Measurement In Tamil Samacheer Kalvi 11th Physics Solutions Chapter 1

(ii) Write a note on triangulation method and radar method to measure larger distances. Triangulation method for the height of an accessible object;
Let AB = h be the height of the tree or tower to be measured. Let C be the point of observation at distance x from B. Place a range finder at C and measure the angle of elevation, ACB =
θ as shown in figure.
From right angled triangle ABC,
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature Of Physical World And Measurement
Samacheer Kalvi 11th Physics Solution Book Chapter 1 Nature Of Physical World And Measurement
(or) height h = x tan θ
Knowing the distance x, the height h can be determined.
RADAR method
The word RADAR stands for radio detection and ranging. A radar can be used to measure accurately the distance of a nearby planet such as Mars. In this method, radio waves are sent from transmitters which, after reflection from the planet, are detected by the receiver. By measuring, the time interval (t) between the instants the radio waves are sent and received, the distance of the planet can be determined as where v is the speed of the radio wave. As the time taken (t) is for the distance covered during the forward and backward path of the radio waves, it is divided by 2 to get the actual distance of the object. This method can also be used to determine the height, at which an aeroplane flies from the ground.
11 Th Physics Samacheer Kalvi Guide Chapter 1 Nature Of Physical World And Measurement

Question 2.
Explain in detail the various types of errors.
Answer:
The uncertainty in a measurement is called an error. Random error, systematic error and gross error are the three possible errors.
(i) Systematic errors: Systematic errors are reproducible inaccuracies that are consistently *, in the same direction. These occur often due to a problem that persists throughout the experiment. Systematic errors can be classified as follows
(1) Instrumental errors: When an instrument is not calibrated properly at the time of manufacture, instrumental errors may arise. If a measurement is made with a meter scale whose end is worn out, the result obtained will have errors. These errors can be corrected by choosing the instrument carefully.
(2) Imperfections in experimental technique or procedure: These errors arise due to the limitations in the experimental arrangement. As an example, while performing experiments with a calorimeter, if there is no proper insulation, there will be radiation losses. This results in errors and to overcome these, necessary correction has to be applied
(3) Personal errors: These errors are due to individuals performing the experiment, may be due to incorrect initial setting up of the experiment or carelessness of the individual making the observation due to improper precautions.
(4) Errors due to external causes: The change in the external conditions during an experiment can cause error in measurement. For example, changes in temperature, humidity, or pressure during measurements may affect the result of the measurement.
(5) Least count error: Least count is the smallest value that can be measured by the measuring instrument, and the error due to this measurement is least count error. The instrument’s
resolution hence is the cause of this error. Least count error can be reduced by using a high precision instrument for the measurement.
(ii) Random errors: Random errors may arise due to random and unpredictable variations in
experimental conditions like pressure, temperature, voltage supply etc. Errors may also be due to personal errors by the observer who performs the experiment. Random errors are sometimes called “chance error”. When different readings are obtained by a person every time he repeats the experiment, personal error occurs. For example, consider the case of the thickness of a wire measured using a screw gauge. The readings taken may be different for different trials. In this case, a large number of measurements are made and then the arithmetic mean is taken.
Samacheer Kalvi 11th Physics Guide Solutions Chapter 1 Nature Of Physical World And Measurement

Question 3.
What do you mean by propagation of errors? Explain the propagation of errors in addition and multiplication.
Answer:
A number of measured quantities may be involved in the final calculation of an experiment. Different types of instruments might have been used for taking readings. Then we may have to look at the errors in measuring various quantities, collectively.
The error in the final result depends on
(i) The errors in the individual measurements
(ii) On the nature of mathematical operations performed to get the final result. So we should know the rules to combine the errors.
The various possibilities of the propagation or combination of errors in different mathematical operations are discussed below:
(i) Error in the sum of two quantities
Let ∆A and ∆B be the absolute errors in the two quantities A and B respectively. Then, Measured value of A = A ± ∆A
Measured value of B = B ± ∆B
Consider the sum, Z = A + B
The error ∆Z in Z is then given by
11th Physics Chapter 1 Numerical Problems Samacheer Kalvi Nature Of Physical World And Measurement
The maximum possible error in the sum of two quantities is equal to the sum of the absolute errors in the individual quantities.
Error in the product of two quantities: Let ∆A and ∆B be the absolute errors in the two quantities A, and B, respectively. Consider the product Z = AB
The error ∆Z in Z is given by Z ± ∆Z = (A ± ∆A) (B ± ∆B)
= (AB) ± (A ∆ B) ± (B ∆ A) ± (∆A • ∆B)
Dividing L.H.S by Z and R.H.S by AB, we get,
11th Physics Samacheer Kalvi Solutions Chapter 1 Nature Of Physical World And Measurement
As ∆A/A, ∆B/B are both small quantities, their product term Samacheer Kalvi Physics 11th Solutions Chapter 1 Nature Of Physical World And Measurement can be neglected.
The maximum fractional error in Z is
Physics Class 11 Samacheer Kalvi Solutions Chapter 1 Nature Of Physical World And Measurement

Question 4.
Write short note on the following:
(a) Unit
(b) Rounding – off
(c) Dimensionless quantities
Answer:
(a) Unit: An arbitrarily chosen standard of measurement of a quantity, which is accepted internationally is called unit of the quantity.
The units in which the fundamental quantities are measured are called fundamental or base units and the units of measurement of all other physical quantities, which can be obtained by a suitable multiplication or division of powers of fundamental units, are called derived units.
(b) Rounding – off: In no case should the result have more significant figures than die figures involved in the data used for calculation. The result of calculation with numbers containing more than one uncertain digit should be rounded off. The rules for rounding off are given below.
Samacheer Kalvi Guru 11 Physics Solutions Chapter 1 Nature Of Physical World And Measurement
(c) Dimensionless quantities: On the basis of dimension, dimensionless quantities are classified into two categories.
(i) Dimensionless variables:
Physical quantities which have no dimensions, but have variable values are called dimensionless variables. Examples are specific gravity, strain, refractive index etc.
(ii) Dimensionless Constant:
Quantities which have constant values and also have no dimensions are called dimensionless constants. Examples are π, e, numbers etc.

Question 5.
Explain the principle of homogeniety of dimensions. What are its uses? Give example.
Answer:
The principle of homogeneity of dimensions states that the dimensions of all the terms in a physical expression should be the same. For example, in the physical expression v2 = u2 + 2as, the dimensions of v2, u2 and 2 as are the same and equal to [L2T-2].
(i) To convert a physical quantity from one system of units to another: This is based on the fact that the product of the numerical values (n) and its corresponding unit (u) is a constant, i.e, = constant n1[u1] = constant (or) n1[u1] = n2[u2].
Consider a physical quantity which has dimension ‘a’ in mass, ‘b’ in length and ‘c’ in time. If the fundamental units in one system are M1, L1 and T1 and the other system are M2, L2 and T2 respectively, then we can write, Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 31
We have thus converted the numerical value of physical quantity from one system of units into the other system.
Example: Convert 76 cm of mercury pressure into Nm-2 using the method of dimensions. Solution: In cgs system 76 cm of mercury pressure = 76 × 13.6 × 980 dyne cm-2
The dimensional formula of pressure P is [ML-1T-2]

(ii) To check the dimensional correctness of a given physical equation:
Example: The equation \(\frac{1}{2} m v^{2}\) = mgh can be checked by using this method as follows.
Solution: Dimensional formula for
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 311
Dimensional formula for
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 32
Both sides are dimensionally the same, hence the equations Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 300 is dimensionally correct.
(iii) To establish the relation among various physical quantities:
Example: An expression for the time period T of a simple pendulum can be obtained by using this method as follows.
Let true period T depend upon
(i) mass m of the bob
(ii) length l of the pendulum and
(iii) acceleration due to gravity g at the place where the pendulum is suspended. Let the constant involved is K = 2π.
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 312
Here k is the dimensionless constant. Rewriting the above equation with dimensions.
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 321
Comparing the powers of M, L and T on both sides, a – 0, b + c = 0, -2c = 1
Solving for a, b and c a = 0, b = 1/2, and c = -1/2
From the above equation
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 33

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Numerical Problems

Question 1.
In a submarine equipped with sonar, the time delay between the generation of a pulse and its echo after reflection from an enemy submarine is observed to be 80 sec. If the speed of sound in water is 1460 ms-1. What is the distance of enemy submarine?
Answer:
Given:
Speed of sound in water = 1460 ms-1
Time delay = 80s
Distance of enemy ship = ?
Solution:
Total distance covered = speed × time
= 1460 ms-1 × 80s = 116800 m
Time taken is for forward and backward path of sound waves.
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 34
= 58400 m (or) 58.4 km

Question 2.
The radius of the circle is 3.12 m. Calculate the area of the circle with regard to significant figures.
Answer:
Given: radius : 3.12 m (Three significant figures)
Solution:
Area of the circle = πr2 = 3.14 × (3.12 m)2 = 30.566
If the result is rounded off into three significant figure, area of the circle = 30.6 m2

Question 3.
Assuming that the frequency v of a vibrating string may depend upon
(i) applied force (F)
(ii) length (l)
(iii) mass per unit length (m), prove that Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 315 using dimensional analysis. [Related to JIPMER 2001]
Answer:
Given: The frequency v of a vibrating string depends
(i) applied force (F)
(ii) length (l)
(iii) mass per unit length (m)
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 351
Substitute the dimensional formulae of the above quantities
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 36
Comparing the powers of M, L, T on both sides,
x + z = 0, x + y – z = 0, -2x = -1
Solving for x, y, z, we get
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 37
Substitute x, y, z values in equ(1)
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 38

Question 4.
Jupiter is at a distance of 824.7 million km from the Earth. Its angular diameter is measured to be 35.72″. Calculate the diameter of Jupiter.
Answer:
Given,
Given Distance of Jupiter = 824.7 × 106 km = 8.247 × 1011 m
angular diameter = 35.72 × 4.85 × 10-6rad = 173.242 × 10-6 rad
= 1.73 × 10-4 rad
∴ Diameter of Jupiter D = D × d = 1.73 × 10-4 rad × 8.247 × 1011 m
= 14.267 × 1o7 m = 1.427 × 108 m (or) 1.427 × 105</sup km

Question 5.
The measurement value of length of a simple pendulum is 20 cm known with 2 mm accuracy. The time for 50 oscillations was measured to be 40 s within 1 s resolution. Calculate the percentage of accuracy in the determination of acceleration due to gravity ‘g’ from the above measurement.
Answer:
Given,
Length of simple pendulum (l) = 20 cm
absolute error in length (∆l) = 2 mm = 0.2 cm
Time taken for 50 oscillation (t) = 40 s
error in time ∆T = 1 s
Solution: Time period for one oscillation (T)
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 39
Hence, the percentage error in g is
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 40

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Conceptual Questions

Question 1.
Why is it convenient to express the distance of stars in terms of light year (or) parsec rather than in km?
Answer:
A parsec is 206, 265 AU and is roughly the distance to the nearest stars. If we were to view a giant star with a diameter of 1 AU at a distance of one parsec, it would appear to be just 1/3600th of a degree in angular size. For comparison, the sun and moon are both half a degree in angular size when viewed from Earth.

Question 2.
Show that a screw gauge of pitch 1 mm and 100 divisions is more precise than a vernier caliper with 20 divisions on the sliding scale.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 41
As shown, the least count of screw gauge is lesser then vernier caliper, hence screw gauge is more precise.

Question 4.
Having all units in atomic standards is more useful. Explain.
Answer:
An atomic mass unit (symbolized AMU or amu) is defined as precisely 1/12 the mass of an atom of carbon-12. The carbon-12 (C-12) atom has six protons and six neutrons in its nucleus.
In imprecise terms, one AMU is the average of the proton rest mass and the neutron rest mass. This is approximately 1.67377 × 10-27 kilogram (kg), or 1.67377 × 10-24 gram (g). The mass of an atom in AMU is roughly equal to the sum of the number of protons and neutrons in the nucleus.
The AMU is used to express the relative masses of, and thereby differentiate between, various isotopes of elements. Thus, for example, uranium-235 (U-235) has an AMU of approximately 235, while uranium-238 (U-238) is slightly more massive. The difference results from the fact that U-238, the most abundant naturally occurring isotope of uranium, has three more neutrons than U-235, an isotope that has been used in nuclear reactors and atomic bombs.

Question 5.
Why dimensional methods are applicable only up to three quantities?
Answer:
Understanding dimensions is of utmost importance as it helps us in studying the nature of physical quantities mathematically. The basic concept of dimensions is that we can add or subtract only those quantities which have same dimensions. Also, two physical quantities are equal if they have same dimensions. these basic ideas help us in deriving the new relation between physical quantities, it is just like units.

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Additional Questions Solved

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Multiple Choose Questions

Question 1.
The unit of surface tension ……
(a) MT-2
(b) Nm-2
(c) Nm
(d) Nm-1
Answer:
(d) Nm-1

Question 2.
One atomus equal to ……
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 401
Answer:
(c) 160 ms

Question 3.
One light year is ……
(a) 3.153 × 107 m
(b) 1.496 × 107 m
(c) 9.46 × 1012 km
(d) 3.26 × 1015 km
Answer:
(c) 9.46 × 1012 km

Question 4.
One Astronomical unit is
(a) 3.153 × 107 m
(b) 1.496 × 107 m
(c) 9.46 × 1012 m
(d) 3.26 × 1015 m
Answer:
(b) 1.496 × 107 m

Question 5.
One parsec is …..
(a) 3.153 × 107 m
(b) 3.26 × 1015 m
(c) 30.84 × 1015 m
(d) 9.46 × 1015 m
Answer:
(c) 30.84 × 1015 m

Question 6.
One Fermi is …..
(a) 10-9 m
(b) 10-10 m
(c) 10-12 m
(d) 10-15 m
Answer:
(d) 10-15 m

Question 7.
One Angstrom is ………
(a) 10-9 m
(b) 10-10m
(c) 10-12 m
(d) 10-15 m
Answer:
(b) 10-10 m

Question 8.
One solar mass is ….
(a) 2 × 1030 kg
(b) 2 × 1030 g
(c) 2 × 1030 mg
(d) 2 × 1030 tonne
Answer:
(d) 2 × 1030 tonne

Question 9.
\(\frac{1}{12}\) of the mass of carbon 12 atom is …..
(a) 1 TMC
(b) mass of neutron
(c) 1 amu
(d) mass of hydrogen
Answer:
(d) mass of hydrogen

Question 10.
The word physics is derived from the word …..
(a) scientist
(b) fusis
(c) fission
(d) fusion
Answer:
(b) fusis

Question 11.
The study of forces acting on bodies whether at rest or in motion is …..
(a) classical mechanics
(b) quantum mechanics
(c) thermodynamics
(d) condensed matter physics
Answer:
(a) classical mechanics

Question 12.
Mass of observable universe …..
(a) 1031 kg
(b) 1041 kg
(c) 1055 kg
(d) 9.11 × 1031 kg
Answer:
(c) 1055 kg

Question 13.
Mass of an electron …….
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 301
Answer:
(b) 9.11 × 10-31 kg

Question 14.
The study of production and propagation of sound waves …..
(a) Astrophysics
(b) Acoustics
(c) Relativity
(d) Atomic physics
Answer:
(b) Acoustics

Question 15.
The study of the discrete nature of phenomena at the atomic and subatomic levels.
(a) Quantum mechanics
(b) High energy physics
(c) Acoustics
(d) Classical mechanics
Answer:
(a) Quantum mechanics

Question 16.
The techniQuestion used to study the crystal structure of various rocks are …….
(a) diffraction
(b) interference
(c) total internal reflection
(d) refraction
Answer:
(a) diffraction

Question 17.
The astronomers used to observe distant points of the universe by …….
(a) Electron telescope
(b) Astronomical telescope
(c) Radio telescope
(d) Radar
Answer:
(c) Radio telescope

Question 18.
The comparison of any physical quantity with its standard unit is known as ……..
(a) fundamental quantities
(b) measurement
(c) dualism
(d) derived quantities
Answer:
(b) measurement

Question 19.
Fundamental quantities can also be known as …… quantities.
(a) original
(b) physical
(c) negative
(d) base
Answer:
(d) base

Question 20.
Which one of the following is not a fundamental quantity?
(a) length
(b) luminous intensity
(c) temperature
(d) water current
Answer:
(d) water current

Question 21.
The system of unit not only based on length, mass, and time is
(a) FPS
(b) CGS
(c) MKS
(d) SI
Answer:
(d) SI

Question 22.
The coherent system of units …..
(a) CGS
(b) SI
(c) FPS
(d) MKS
Answer
(b) SI

Question 23.
The triple point temperature of water is ……
(a) -273.16 K
(b) 0K
(c) 273.16 K
(d) 100 K
Answer:
(d) 100 K

Question 24.
Which of the following is a unit of distance?
(a) Light year
(b) Leap year
(c) Dyne-sec
(d) Pauli
Answer:
(a) Light year

Question 25.
The unit of moment of force ……
(a) Nm2
(b) Nm
(c) N
(d) NJ rad
Answer:
(b) Nm

Question 26.
1 radian is ……
(a) 2.91 × 10-4 m
(b) 57.27°
(c) 180°
(d) \(\frac{\pi}{180}\)
Answer:
(b) 57.27°

Question 27.
One degree of arc is …….
(a) 1″
(b) 60″
(c) 60′
(d) 60°
Answer:
(c) 60′

Question 28.
One degree of arc is equal to …….
(a) 1.457 × 102 rad
(b) 1.457 × 10-2 rad
(c) 1.745 × 102 rad
(d) 1.745 × 10-2 rad
Answer:
(b) 1.457 × 10-2 rad

Question 29.
1 minute of arc is equal to …….
(a) 1.745 × 10-2 rad
(b) 2.91 × 10-4 rad
(c) 2.91 × 104 rad
(d) 4.85 × 10-6 rad
Answer:
(b) 2.91 × 10-4 rad

Question 30.
1 second of arc is equal to ………
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 35
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 361

Question 31.
1 second of arc is equal to ….
(a) 0.00027°
(b) 1.745 × 10-2 rad
(c) 2.91 × 10-4 rad
(d) 4.85 × 10-6 rad
Answer:
(a) 0.00027°

Question 32.
Unit of impulse ….
(a) NS2
(b) NS
(c) Nm
(d) Kgms-2
Answer:
(b) NS

Question 33.
The ratio of energy and temperature is known as ……
(a) Stefen’s constant
(b) Boltzmann constant
(c) Plank’s constant
(d) Kinetic constant
Answer:
(b) Boltzmann constant

Question 34.
The range of distance can be measured by using direct methods is …..
(a) 10-2 to 10-5 m
(b) 10-2 to 102 m
(c) 102 to 1(T5 m {d) 10″2 to 105 m
Answer:
(b) 10-2 to 102 m

Question 35.
Which of the following is in increased order?
(a) exa, tera, hecto
(b) tera, exa, hecto
(c) giga, tera, exa
(d) hecto, exa, giga
Answer:
(c) giga, tera, exa

Question 36.
10-18 is called as ……
(a) nano
(b) pico
(c) femto
(d) atto
Answer:
(d) atto

Question 37.
A radio signal sent towards the distant planet, returns after “t” s. If “c” is the speed of radio
waves then the distance of the planet and from the earth is …….
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 50
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 51

Question 38.
Find odd one out ….
(a) Newton
(b) metre
(c) candela
(d) Kelvin
Answer:
(a) Newton

Question 39.
The shift in the position of an object when viewed with two eyes, keeping one eye closed at a
time is known as …
(a) basis
(b) fundamental
(c) parallax
(d) pendulum
Answer:
(c) parallax

Question 40.
Chandrasekar limit is ….. times the mass of the sun.
(a) 1.2
(b) 1.4
(c) 1.6
(d) 1.8
Answer:
(b) 1.4

Question 41.
The smallest physical unit of time is
(a) second
(b) minute
(c) microsecond
(d) shake
Answer:
(d) shake

Question 42.
Size of atomic nucleus is …..
(a) 10-10 m
(b) 10-12 m
(c) 10-14 m
(d) 10-18 m
Answer:
(c) 10-14 m

Question 43.
Time interval between two successive heart beat is in the order of …….
(a) 10° s
(b) 10 s
(c) 102 s
(d) 10-3 s
Answer:
(a) 10° s

Question 44.
Half life time of a free neutron is in the order of ……
(a) 10°
(b) 101 s
(c) 102 s
(d) 103 s
Answer:
(d) 103 s

Question 45.
The uncertainty contained in any measurement is ……
(a) rounding off
(b) error
(c) parallax
(d) gross
Answer:
(b) error

Question 46.
Zero error of an instrument is a ……
(a) Systematic error
(b) Random error
(c) Gross error
(d) Both (a) and (b)
Answer:
(a) Systematic error

Question 47.
Error in the measurement of radius of a sphere is 2%. Then error in the measurement of surface
area is ….
(a) 1%
(b) 2%
(c) 3%
(d) 4%
Answer:
(d) 4%

Question 48.
Imperfections in experimental procedure gives ….. error.
(a) random
(b) gross
(c) systematic
(d) personal
Answer:
(c) Systematic

Question 49.
Random error can also be called as ….
(a) personal error
(b) chance error
(c) gross error
(d) system error
Answer:
(b) chance error

Question 50.
To get the best possible true value of the quantity has to be taken.
(a) rms value
(b) net value
(c) arithmetic mean
(d) mode
Answer:
(c) arithmetic mean

Question 51.
The error caused due to the shear carelessness of an observer is called as …… error.
(a) Systematise
(b) Gross
(c) Random
(d) Personal
Answer:
(b) Gross

Question 52.
The uncertainty in a measurement is called as ….
(a) error
(b) systematic
(c) random error
(d) gross error
Answer:
(a) error

Question 53.
The difference between the true value and the measured value of a quantity is known as …..
(a) Absolute error
(b) Relative error
(c) Percentage error
(d) Systemmatic error
Answer:
(a) Absolute error

Question 54.
If a1, a2, a3 …. an are the measured value of a physical quantity “a” and am is the true value then absolute error …..
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 60
Answer:
(d) \(\Delta a_{n}=a_{m}-a_{n}\)

Question 55.
If ‘am‘ and ‘∆am ‘ are true value and mean absolute error respectively, then the magnitude of the quantity may lie between …..
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 61
Answer:
(b) \(a_{m}-\Delta a_{m} \text { to } a_{m}+\Delta a_{m}\)

Question 56.
The ratio of the mean absolute error to the mean value is called as ……
(a) absolute error
(b) random error
(c) relative error
(d) percentage error
Answer:
(c) Relative error

Question 57.
Relative error can also be called as ……
(a) fractional error
(b) absolute error
(c) percentage error
(d) systematic error
Answer:
(a) fractional error

Question 58.
A measured value to be close to targeted value, percentage error must be close to
(a) 0
(b) 10
(c) 100
(d) ∝
Answer:
(a) 0

Question 59.
The maximum possible error in the sum of two quantities is equal to …….
(a) Z = A + B
(b) ∆Z = ∆A + ∆B
(c) ∆Z = ∆A/∆B
(d) ∆Z = ∆A – ∆B
Answer:
(b) ∆Z = ∆A + ∆B

Question 60.
The maximum possible error in the difference of two quantities is ……
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 62
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 63

Question 61.
The maximum fractional error in the division of two quantities is ….
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 64
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 65

Question 62.
The fractional error in the nth power of a quantity is …..
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 66
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 67

Question 63.
A physical quantity is given as y = \(\frac{a b^{3}}{c^{2}}\). If ∆a, ∆b, ∆c are absolute errors, the possible fractional error in y is …..
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 68
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 69

Question 64.
Number of significant digits in 3256 …
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 65.
Number of significant digits in 32005 ……
(a) 1
(b) 2
(c) 5
(d) 2
Answer:
(c) 5

Question 66.
Number of significant digits in 2030 …
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 67.
Number of significant digits in 2030N …..
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 68.
Number of significant digits in 0.0342 …..
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c)3

Question 69.
Number of significant digit in 20.00 …..
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 70.
Number of significant digit in 0.030400
(a) 6
(b) 5
(c) 4
(d) 3
Answer:
(b) 5

Question 71.
The force acting on a body is measured as 4.25 N. Round it off with two significant figure ..
(a) 4.3
(b) 4.2
(c) both
(a) or (b)
(d) 4.25
Answer:
(b) 4.2

Question 72.
The quantities a, b, c are measured as 3.21, 4.253, 7.2346. The sum (a + b + c) with proper
significant digits is ……
(a) 14.6976
(b) 14.697
(c) 14.69
(d) 14.6
Answer:
(c) 14.69

Question 73.
The dimensions of gravitational constant G are …
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 80
Answer:
(b) \(\mathbf{M}^{-1} \mathbf{L}^{3} \mathbf{T}^{-2}\)

Question 74.
The ratio of one nanometer to one micron is
(a) 10-3
(b) 103
(c) 10-9
(d) 10-6
Answer:
(b) 103

Question 75.
Which of the following pairs does not have same dimension?
(a) Moment of inertia and moment of force
(b) Work and torque
(c) Impulse and momentum
(d) Angular momentum and Plank’s constant
Answer:
(a) Moment of inertia and moment of force

Question 76.
Two quantities A and B have different dimensions. Which of the following is physically meaningful?
(a) A + B
(b) A – B
(c) A /B
(d) None
Answer:
(c) A /B

Question 77.
The dimensional formula for moment of inertia ……
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 81
Answer:
(d) \(\mathbf{M} \mathbf{L}^{2} \mathbf{T}^{\mathbf{0}}\)

Question 78.
Which of the following is having same dimensional formula?
(a) Work and power
(b) Radius of gyration and displacement
(c) Impulse and force
(d) Frequencies and wavelength
Answer:
(b) Radius of gyration and displacement

Question 79.
Which of the following quantities is expressed as force per unit area?
(a) Pressure
(b) Stress
(c) Both (a) and (b)
(d) None
Answer:
(c) Both (a) and (b)

Question 80.
In equation of motion Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 85 the dimensional formula for K is …..
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 86
Answer:
(b) \(\left[\mathbf{L} \mathbf{T}^{-2}\right]\)

Question 81.
The dimensional formula for heat capacity ……
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 87
Answer:
(d) \(\left[\mathbf{M} \mathbf{L}^{2} \mathbf{T}^{2} \mathbf{K}^{-1}\right]\)

Question 82.
The product of Avogadro constant and elementary charge is known as …… constant.
(a) Planck’s
(b) Avagadro
(c) Boltzmann
(d) Faraday
Answer:
(d) Faraday

Question 83.
The force F is given by F = at + bt2 where t is time. The dimensions of ‘a’ and ‘b’ respectively are
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 861
Answer:
(d) \(\left[\mathrm{MLT}^{-2}\right]\) and \(\left[\mathrm{MLT}^{-0}\right]\)

Question 84.
Dimensions of impulse are …..
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 871
Answer:
(c) \(\left[\mathrm{MLT}^{-1}\right]\)

Question 85.
If speed of light (c), acceleration due to gravity (g) and pressure (P) are taken as fundamental
units, the possible relation to gravitational constant (G) is ….
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 88
Answer:
(c) \(c^{0} g^{2} p^{-1}\)

Question 86.
Equivalent of one joule is ……
(a) Nm2
(b) kg m2 s-2
(c) kg m s-1
(d) N kg m2
Answer:
(b) kg m2 s-2

Question 87.
Pick out the dimensionless quantity …..
(a) force
(b) specific gravity
(c) planck’s constant
(d) velocity
Answer:
(b) specific gravity

Question 88.
Odd one out ………
(a) strain
(b) refractive index
(c) numbers
(d) stress
Answer:
(d) stress

Question 89.
A wire has a mass 0.3 ± 0.003g, radius 0.5 ± 0.005 mm and length 6 + 0.06 cm. The maximum percentage error in the measurement of its density is …….
(a) 1%
(b) 2%
(c) 3%
(d) 4%
Answer:
(d) 4%

Question 90.
The dimensions of planck constant equals to that of …..
(a) energy
(b) momentum
(c) angular momentum
(d) power
Answer:
(c) angular momentum

Question 91
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 90
(a) The unit of λ is same as that of x and A
(b) The unit of λ is same as that of x but not of A
(c) The unit of c is same as that of 2π/λ
(d) The unit of (ct – x)is same as that 2π/λ
Answer:
(a) The unit of λ is same as that of x and A

Question 92.
The number of significant figures in 0.06900 is …….
(a) 2
(b) 2
(c) 4
(d) 5
Answer:
(c) 4

Question 93.
The numbers 3.665 and 3.635 on rounding off to 3 significant figures will give
(a) 3.66 and 3.63
(b) 3.66 and 3.64
(c) 3.67 and 3.63
(d) 3.67 and 3.64
Answer:
(b) 3.66 and 3.64

Question 94.
Which of the following measurements is most precise?
(a) 4.00 mm
(b) 4.00 cm
(c) 4.00 m
(d) 4.00 km
Answer:
(a) 4.00 mm

Question 95.
The mean radius of a wire is 2 mm. Which of the following measurements is most accurate? (a) 1.9 mm
(b) 2.25 mm
(c) 2.3 mm
(d) 1.83 mm
Answer:
(a) 1.9 mm

Question 96.
If error in measurement of radius of sphere is 1%. What will be the error in measurement of volume?
(a) 1%
(b) \(\frac{1}{3}\)%
(c) 3%
(d) 10%
Answer:
(c) 3%

Question 97.
Dimensions [M L-1 T-1] are related to …….
(a) torque
(b) work
(c) energy
(d) Coefficient of viscosity
Answer:
(d) Coefficient of viscosity

Question 98.
Heat produced by a current is obtained a relation H = I2RT. If the errors in measuring these quantities current, resistance, time are 1%, 2%, 1% respectively then total error in calculating the energy produced is
(a) 2%
(b) 4%
(c) 5%
(d) 6%
Answer:
(c) 5%

Question 99.
Length cannot be measured by ….
(a) fermi
(b) angstrom
(c) parsec
(d) debye
Answer:
(d) debye

Question 100.
The pressure on a square plate is measured by measuring the force on the plate and the length of the sides of the plate by using the formula p = \([/late\frac{\mathrm{F}}{l^{2}}x]. If the maximum errors in the measurement
of force and length are 4% and 2% respectively, then the maximum error in the measurement of pressure is ……..
(a) 1%
(b) 2%
(c) 8%
(d) 10%
Aswer:
(c) 8%

Question 101.
Which of the following cannot be verified by using dimensional analysis?
1 mv2
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 100
Answer:
(b) y = a sin wt

Question 102.
Percentage errors in the measurement of mass and speed are 3% and 2% respectively. The error in the calculation of kinetic energy is …….
(a) 2%
(b) 3%
(c) 5%
(d) 7%
Answer:
(d) 7%

Question 103.
More number of readings will reduce …….
(a) random error
(b) systematic error
(c) both (a) and (b)
(d) neither (a) nor (b)
Answer:
(a) random error

Question 104.
If the percentage error in the measurement of mass and momentum of a body are 3% and
2%respectively, then maximum possible error in kinetic energy is
(a) 2%
(b) 3%
(c) 5%
(d) 7%
Answer:
(d) 7%

Question 105.
In a vernier caliper, n divisions of vernier scale coincides with (n – 1) divisions of main scale. The least count of the instrument is ………
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 121
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 103

Question 106.
The period of a simple pendulum is recorded as 2.56s, 2.42s, 2.71s and 2.80s respectively.
The average absolute error is
(a) 0.1s
(b) 0.2s
(c) 1.0s
(d) 0.11s
Answer:
(d) 0.11s

Question 107.
In a system of units, if force (F), acceleration (A) and time (T) are taken as fundamental units
then the dimensional formula of energy is
(a) [FA2T]
(b) [FAT2]
(c) [F2AT]
(d) [FAT]
Answer:
(b) [FAT2]

Question 108.
The random error in the arithmetic mean of 50 observations is ‘a’, then the random error in the
arithmetic mean of 200 observations a would be
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 104
Answer:
(c) [latex]\frac{a}{4}\)

Question 109.
Which of the following is not dimensionless?
(a) Relative permittivity
(b) Refractive index
(c) Relative density
(d) Relative velocity
Answer:
(d) Relative velocity

Question 110.
If V-velocity, K – kinetic energy and T – time are chosen as the fundamental units, then what is the dimensional formula for surface tension?
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 110
Answer:
(a) [K V-2 T-2]

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Short Answer Questions (1 Mark)

Question 1.
A new unit of length is chosen such that the speed of light in vaccum is unity. What is the distance between the sun and the earth in terms of the new unit if light takes 8 min and 20 s to cover this distance.
Answer:
Speed of light in vacuum, c = 1 new unit of length s-1
t = 8 min. 20 sec, = 500 s
x = ct= 1 new unit of length s-1 × 500s
x = 500 new unit of length

Question 2.
If x = a + bt + ct2, where x is in metre and t in seconds, what is the unit of c ?
Answer:
The unit of left hand side is metre so the units of ct2 should also be metre.
Since t2 has unit of s2, so the unit of c is m/s2.

Question 3.
What is the difference between mN, Nm and nm ?
Answer:
mN means milli newton, 1 mN = 10-3 N, Nm means Newton meter, nm means nano meter.

Question 4.
The radius of atom is of the order of 1 A° & radius of nucleus is of the order of fermi. How many magnitudes higher is the volume of the atom as compared to the volume of nucleus ?
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 152

Question 5.
How many kg make 1 unified atomic mass unit ?
Answer:
1u = 1.66 × 10-27 kg

Question 6.
Name some physical quantities that have same dimension.
Answer:
Work, energy and torque.

Question 7.
Name the physical quantities that have dimensional formula [ML -1T-2].
Answer:
Stress, pressure, modulus of elasticity.

Question 8.
Give two examples of dimensionless variables.
Answer:
Strain, refractive index.

Question 9.
State the number of significant figures in
(i) 0.007 m2
(ii) 2.64 × 1024 kg
(iii) 0.2370 g cm-3
(iv) 0.2300m
(v) 86400
(vi) 86400 m
Answer:
(i) 1,
(ii) 3,
(iii) 4,
(iv) 4,
(v) 3,
(vi) 5 since it comes from a measurement the last two zeros become significant.

Question 10.
Given relative error in the measurement of length is 0.02, what is the percentage error ?
Answer:
2%.

Question 11.
A physical quantity P is related to four observables a, b, c and d as follows :
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 1101
The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2% respectively. What is the percentage error in the quantity P?
Answer:
Relative error in P is given by
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 120

Question 12.
A boy recalls the relation for relativistic mass (m) in terms of rest mass (m0) velocity of particle V, but forgets to put the constant c (velocity of light). He writes Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 125 correct the equation by putting the missing ‘c’.
Answer:
Since quantities of similar nature can only be added or subtracted, v2 cannot be subtracted from 1 but v2/c2 can be subtracted from 1.
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 1212

Question 13.
Name the technique used in locating.
(a) an under water obstacle
(b) position of an aeroplane in space.
Answer:
(a) SONAR ➝ Sound Navigation and Ranging.
(b) RADAR ➝ Radio Detection and Ranging.

Question 14.
Deduce dimensional formulae of—
(i) Boltzmann’s constant
(ii) mechanical equivalent of heat.
Answer:
(i) Boltzmann Constant:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 136

Question 15.
Give examples of dimensional constants and dimensionless constants.
Answer:
Dimensional Constants : Gravitational constant, Plank’s constant. Dimensionless Constants : it, e.

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Short Answer Questions (2 Marks)

Question 16.
The vernier scale of a travelling microscope has 50 divisions which coincide with 49 main scale divisions. If each main scale division is 0.5 mm. Calculate the minimum inaccuracy in the measurement of distance.
Answer:
Minimum inaccuracy = Vernier constant
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 137

Question 17.
If the unit of force is 100N, unit of length is 10m and unit of time is 100s. What is the unit of Mass in this system of units ?
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 138

Question 18.
State the principle of homogeneity. Test the dimensional homogeneity of equations
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 139
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 140
as Dimensions of L.H.S. = Dimensions of R.H.S.
∴ The equation to dimensionally homogeneous.

(ii) Sn = Distance travelled in nth sec that is (Sn – Sn – 1)
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 141
Hence this is dimensionally correct.

Question 19.
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 142
Answer:
Since dimensionally similar quantities can only be added
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 143

Question 20.
Magnitude of force experienced by an object moving with speed v is given by F = kv2. Find dimensions of k.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 144

Question 21.
A book with printing error contains four different formulae for displacement. Choose the
correct formula/formulae
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 145
Answer:
The arguments of sine and cosine function must be dimensionless so (a) is the probable correct formulae. Since
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 146

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Numericals Questions

Question 22.
Determine the number of light years in one metre.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 240

Question 23.
The mass of a box measured by a grocer’s balance is 2.3 kg. Two gold pieces 20.15 g and 20.17 g are added to the box.
(i) What is the total mass of the box ?
(ii) The difference in masses of the pieces to correct significant figures.
Answer:
(i) Mass of box = 2.3 kg
Mass of gold pieces = 20.15 + 20.17 = 40.32 g = 0.04032 kg.
Total mass = 2.3 + 0.04032 = 2.34032 kg
In correct significant figure mass = 2.3 kg (as least decimal)
(ii) Difference in mass of gold pieces = 0.02 g
In correct significant figure (2 significant fig. minimum decimal) will be 0.02 g.

Question 24.
5.74 g of a substance occupies 1.2 cm3. Express its density to correct significant figures.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 245
Here least significant figure is 2, so density = 4.8 g/cm3.

Question 25.
If displacement of a body s = (200 ± 5) m and time taken by it t = (20 + 0.2) s, then find the percentage error in the calculation of velocity.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 246

Question 26.
If the error in measurement of mass of a body be 3% and in the measurement of velocity be 2%. What will be maximum possible error in calculation of kinetic energy.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 247

Question 27.
The length of a rod as measured in an experiment was found to be 2.48 m, 2.46 m, 2.49 m, 2.50 m and 2.48 m. Find the average length, absolute error and percentage error. Express the result with error limit.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 248

Question 28.
A physical quantity is measured as Q = (2.1 ± 0.5) units. Calculate the percentage error in (1) Q2 (2) 2Q.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 249

Question 29.
When the planet Jupiter is at a distance of 824.7 million km from the earth, its angular diameter is measured to be 35.72″ of arc. Calculate diameter of Jupiter.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 250

Question 30.
A laser light beamed at the moon takes 2.56s and to return after reflection at the moon’s surface. What will be the radius of lunar orbit?
Answer:
t = 2.54 s
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 251

Question 31.
Convert
(i) 3 m.s-2 to km h-2
(ii) G = 6.67 × 10-11 N m2 kg-2 to cm3 g-1 s-2
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 252
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 253

Question 32.
A calorie is a unit of heat or energy and it equals 4.2 J where 1J = 1 kg m2s-2. Suppose we employ a system of units in which unit of mass is α kg, unit of length is β m, unit of time γs. What will be magnitude of calorie in terms of this new system.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 254

Question 33.
The escape velocity v of a body depends on—
(i) the acceleration due to gravity ‘g’ of the planet,
(ii) the radius R of the planet. Establish dimensionally the relation for the escape velocity.
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 255

Question 34.
The frequency of vibration of a string depends of on,
(i) tension in the string
(ii) mass per unit length of string,
(iii) vibrating length of the string. Establish dimensionally the relation for frequency.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 256
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 257

Question 35.
One mole of an ideal gas at STP occupies 22.4 L. What is the ratio of molar volume to atomic volume of a mole of hydrogen? Why is the ratio so large? Take radius of hydrogen molecule to be 1°A.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 258
This ratio is large because actual size of gas molecule is negligible in comparison to the inter molecular separation.

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Samacheer Kalvi 11th Computer Science Solutions Chapter 4 Theoretical Concepts of Operating System

Students can Download Computer Science Chapter 4 Theoretical Concepts of Operating System Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Computer Science Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Computer Science Solutions Chapter 4 Theoretical Concepts of Operating System

Samacheer Kalvi 11th Computer Science Theoretical Concepts of Operating System Text Book Back Questions and Answers

PART – 1
I. Choose The Correct Answer

11th Computer Science Chapter 4 Book Back Answers Question 1.
Operating system is a ……………….
(a) Application Software
(b) Hardware
(c) System Software
(d) Component
Answer:
(c) System Software

Theoretical Concepts Of Operating System Question 2.
Identify the usage of Operating Systems ……………….
(a) Easy interaction between the human and computer
(b) Controlling input & output Devices
(c) Managing use of main memory
(d) All the above
Answer:
(d) All the above

Computer Operating System In Tamil Pdf Question 3.
Which of the following is not a function of an Operating System?
(a) Process Management
(b) Memory Management
(c) Security management
(d) Compiler Environment
Answer:
(d) Compiler Environment

Samacheer Kalvi Advantages And Disadvantages Question 4.
Which of the following OS is a Commercially licensed Operating system?
(a) Windows
(b) UBUNTU
(c) FEDORA
(d) REDHAT
Answer:
(a) Windows

Samacheer Kalvi Guru 11th Computer Science Question 5.
Which of the following Operating systems support Mobile Devices?
(a) Windows 7
(b) Linux
(c) BOSS
(d) iOS
Answer:
(d) iOS

Question 6.
File Management manages ……………….
(a) Files
(b) Folders
(c) Directory systems
(d) All the Above
Answer:
(d) All the Above

Question 7.
Interactive Operating System provides ……………….
(a) Graphics User Interface (GUI)
(b) Data Distribution
(c) Security Management
(d) Real Time Processing
Answer:
(a) Graphics User Interface (GUI)

Question 8.
Android is a ……………….
(a) Mobile Operating system
(b) open Source
(c) Developed by Google
(d) All the above
Answer:
(d) All the above

Question 9.
Which of the following refers to Android operating system’s version?
(a) JELLY BEAN
(b) UBUNTU
(c) OS/2
(d) MITTIKA
Answer:
(a) JELLY BEAN

PART – 2
II. Short Answers

Question 1.
What are the advantages of memory management in Operating System?
Answer:

  1. Allocating memory is easy and cheap
  2. Any free page is ok, OS can take first one out of.list it keeps
  3. Eliminates external fragmentation
  4. Data (page frames) can be scattered all over PM
  5. Pages are mapped appropriately anyway
  6. Allows demand paging and pre – paging
  7. More efficient swapping
  8. No need for considerations about fragmentation
  9. Just swap out page least likely to be used

Question 2.
What is the multi – user Operating system?
Answer:
A Multi – user Operating system is a computer operating system (OS) that allows multiple users on different computers or terminals to access a single system with one OS on it.

Question 3.
What is a GUI?
Answer:
GUI – Graphical User Interface – It allows the use of icons or other visual indicators to interact with electronic devices, rather than using only text via the command line. For example, all versions of Microsoft Windows utilize a GUI, whereas MS – DOS does not.

Question 4.
List out different distributions of Linux operating system.
Answer:
Different server distributions in Linux OS:

  1. Ubuntu
  2. Linux Mint
  3. Debian
  4. Fedora
  5. RedHa.

Question 5.
What are the security management features available in Operating System?
Answer:
Security Management features:

  1. File access level security
  2. System level security
  3. Network level securny.

Question 6.
What is multi – processing?
Answer:
Multiprocessing is one of the features of operating system. It has two or more processors for a single running process. Each processor works on different parts of the same task or two or more different tasks.

Question 7.
What are the different Operating Systems used in computer?
Answer:
Different operating system used:

  1. Single user, single Task Operating system
  2. Multi user operating system
  3. Multiprocessing operating system
  4. Distributed Operating system

PART – 3
III. Explain in Brief

Question 1.
What are the advantages and disadvantages of Time – sharing features?
Answer:
Advantages and disadvantages of Time Sharing Option (TSO)
Advantages:

  • Each task and each user get its time.
  • Systems have to give time to these application individual tasks and other applications also, so that all system behave correctly.
  • Reduces the CPU ideal time

Disadvantages:

  • Problem in Reliability
  • It consumes many resources so it need special operating systems.
  • Need High specification hardware

Question 2.
Explain and List out examples of mobile operating system.
Answer:
A Mobile Operating System (or mobile OS) is an operating system that is specifically designed to run on mobile devices such as phones, tablets, smart watches, etc.
Example: Apple IOS

Google android : Android is a mobile OS developed by Google, based on Linux and designed for smart phones and tabs. iOS was developed by Apple.
Example : Android, ColorOS, LGUX, MIUI.

Question 3.
What are the differences between Windows and Linux Operating system?
Answer:
Windows:

  • Microsoft windows is a proprietary OS which is commercial.
  • Windows can be modified only by the company that owns it.
  • Difficult to customize.
  • Vulnerable to virus and malware attacks.

Linux:

  • Linux is open source, i.e., free licensed.
  • Linux can be modified by anyone.
  • Easy to customize.
  • More secure.

Question 4.
Explain the process management algorithms in Operating System.
Answer:
The following algorithms are mainly used to allocate the job (process) to the processor: FIFO, SJF, Round Robin, Based on priority.

  1. First In First Out (FIFO) Scheduling – This algorithm is based on queuing technique. Technically, the process that enters the queue first is executed first by the CPU, followed by the next and so on. The processes are executed in the order of the queue.
  2. Shortest Job First (SJF) Scheduling – This algorithm works based on the size of the job being executed by the CPU.
  3. Round Robin Scheduling – Round Robin (RR) Scheduling algorithm is designed especially -for time sharing systems, jobs are assigned and processor time in a circular method.
  4. Based on priority – The given job (process) is assigned on a priority. The job which has higher priority is more important than ether jobs.

PART – 4
IV. Explain in Detail

Question 1.
Explain the concept of a Distributed Operating System.
Answer:
The data and application that are stored and processed on multiple physical locations across the world over the digital network (intemet/intranet). The Distributed Operating System is used to access shared data and files that reside in any machine around the world. The user can handle the data from different locations. The users can access as if it is available on their own computer.
11th Computer Science Chapter 4 Book Back Answers Samacheer Kalvi Theoretical Concepts Of Operating System
The advantages of distributed Operating System are as follows:

  1. A user at one location can make use of all the resources available at another location over the network.
  2. Many computer resources can be added easily in the network
  3. Improves the interaction with the customers and clients.
  4. Reduces the load on the host computer.

Question 2.
Explain the main purpose of an operating system.
Operating system has become essential to enable the users to design applications without the knowledge of the computer’s internal structure of hardware.
Operating system manages all the software and hardware. Most of the time there are many different computer programmes running at the same time, they all need to access the computers, CPU, memory and storage. The need of operating system is basically – an interface between the user and the hardware.
Theoretical Concepts Of Operating System Samacheer Kalvi 11th Computer Science Solutions Chapter 4
Operating System works as translator, while it translates the user request into machine language(Binary language), processes it and then sends it back to Operating System. Operating System converts processed information into user readable form.

Uses of Operating Systems:

  1. to ensure that a computer can be used to extract what the user wants it do.
  2. Easy interaction between the users and computers.
  3. Starting computer operation automatically when power is turned on (Booting).
  4. Controlling Input and Output Devices
  5. Manage the utilisation of main memory.
  6. Providing security to user programs.

Question 3.
Explain advantages and disadvantages of open source operating systems.
Answer:
The benefits of open source are tremendous and have gained huge popularity in the IT field in recent years. They are as follows:

  1. Open source (OS) is free to use, distribute and modify.
  2. Open source is independent of the company as author who originally created it.
  3. It is accessible to everyone. Anyone can debug the coding.
  4. It doesn’t have the problem of incompatible formats that exits in proprietary software.
  5. It is easy to customize as per our needs.
  6. Excellent support will be provided by programmers who will assist in making solutions.

Some of the disadvantages are:

  1. Latest hardware are incompatible, i.e. lack of device drivers.
  2. It is less user friendly and not as easy to use.
  3. There may be some indirect costs involved such as paying for external support.
  4. Malicious users can potentially view it and exploit any vulnerability.

Samacheer kalvi 11th Computer Science Theoretical Concepts of Operating System Additional Questions and Answers

PART – 1
I. Choose the correct answer

Question 1.
Software is classified into ………………. types.
(a) five
(b) two
(c) four
(d) six
Answer:
(b) two

Question 2.
A computer consists of a collection of processes, they are classified as ………………. categories.
(a) 7
(b) 3
(c) 8
(d) 2
Answer:
(d) 2

Question 3.
Which one of the following is not an algorithm?
(a) NTFS
(b) FIFO
(c) SJE
(d) Round Robin
Answer:
(a) NTFS

Question 4.
The operating system provides ………………. levels of securities to the user end.
(a) three
(b) five
(c) seven
(d) ten
Answer:
(a) three

Question 5.
Which one of the following is not a prominent operating system?
(a) UNIX
(b) IOS
(c) GUI
(d) Android
Answer:
(c) GUI

Question 6.
………………. is a family of multitasking.
(a) LINUX
(b) Microsoft Windows
(c) UNIX
(d) iOS
Answer:
(a) LINUX

Question 7.
Which one of the following comes under proprietary licence?
(a) Apple Mac OS
(b) Google’s Android
(c) UNIX
(d) LINUX
Answer:
(a) Apple Mac OS

Question 8.
The LINUX operating system was originated in ……………….
(a) 1996
(b) 1998
(c) 2000
(d) 1991
Answer:
(d) 1991

Question 9.
………………. is the second most popular mobile operating system globally after Android.
(a) Microsoft Windows
(b) iOS
(c) UNIX
(d) LINUX
Answer:
(b) iOS

Question 10.
Which one of the following is an application software to play audio and video files?
(a) Audio Player
(b) Media Player
(c) VLC Player
(d) All of these
Answer:
(c) VLC Player

Question 11.
Which one of the following is a System software?
(a) Operating System
(b) Language Processor
(c) Both a & b
(d) none of these
Answer:
(c) Both a & b

Question 12.
Which one of the following is a set of instructions that perform specific tasks?
(a) Hardware
(b) Software
(c) Processor
(d) I/O devices
Answer:
(b) Software

Question 13.
Hardware and software are managed by ……………….
(a) GUI
(b) OS
(c) Bootstrap
(d) keyboard
Answer:
(b) OS

Question 14.
The process of starting computer operation automatically when the power is turned on is called ……………….
(a) Booting
(b) Compiling
(c) executing
(d) Storing
Answer:
(a) Booting

Question 15.
An OS that allows only a single user to perform a task at a time is called as ……………….
(a) Single user os
(b) Single task os
(c) Both a & b
(d) Multi tasking os
Answer:
(c) Both a & b

Question 16.
Identify the single user and single task OS?
(a) MS – DOS
(b) UNIX
(c) LINUX
(d) iOS
Answer:
(a) MS – DOS

Question 17.
Identify the multi-user OS?
(a) Windows
(b) Linux
(c) UNIX
(d) All of these
Answer:
(d) All of these

Question 18.
To build a cheap computer, ………………. os is used.
(a) Windows
(b) Raspbion OS
(c) iOS
(d) None of these
Answer:
(b) Raspbion OS

Question 19.
GUI stands for ……………….
(a) Geo User Interact
(b) Global User Inter Change
(c) Graphical User Interface
(d) Global User Interface
Answer:
(c) Graphical User Interface

Question 20.
A ………………. is the unit of work or program in a computer.
(a) Process
(b) Code
(c) Concept
(d) Log file
Answer:
(a) Process

Question 21.
The operating system processes are executed by ……………….
(a) User code
(b) System code
(c) Task
(d) Program
Answer:
(b) System code

Question 22.
System level security is provided by ………………. in a multi user environment.
(a) Permission
(b) execute
(c) Password
(d) Security code
Answer:
(c) Password

Question 23.
NTFS is a ……………….
(a) game
(b) file management technique
(c) os
(d) System level security
Answer:
(b) file management technique

Question 24
………………. os is used to access shared data that resides in any machine around the world.
(a) Time sharing
(b) fixed
(c) MS – Dos
(d) distributed
Answer:
(d) distributed

Question 25.
Unix was developed in the year ……………….
(a) 1970
(b) 1980
(c) 1990
(d) 1960
Answer:
(a) 1970

Question 26.
Unix was developed by?
(a) Ken Thompson
(b) Dennis Ritchie
(c) Both a & b
(d) Ricki Mascitti
Answer:
(c) Both a & b

Question 27.
………………. is a windows alternative open source operating system.
(a) React OS
(b) Boss
(c) Redhat
(d) Fedora
Answer:
(a) React OS

Question 28.
Google has developed ………………. for wrist watches.
(a) Android wear
(b) Android wrist
(c) Android wrist watches
(d) Android watches
Answer:
(a) Android wear

Question 29.
Which among the following is not an android moblie open source versions?
(a) Dotnut
(b) Froyo
(c) Nougat
(d) Alpha
Answer:
(a) Dotnut

PART – 2
II. Short Answers

Question 1.
What is an operating system?
Answer:
An operating system is a software which serves as the interface between a user and a computer.

Question 2.
What are the different types of Operating System?
Answer:
Single user and single task operating system, Multi user operating system, Multi – Processing Operating system, Time sharing Operating system.

Question 3.
What is Real Time operating system?
Answer:
It is a multi – tasking and multi user operating system designed for real time based applications such as robotics, weather and climate prediction software etc.,

Question 4.
What is meant by Distributed Operating system?
Answer:
The data and applications are stored and processed on multiple locations in an around world over the digital network.

Question 5.
What are the advantages of Distributed Operating system?
Answer:
Resources can be used in different locations. Improves interaction with customers and clients. Reduces load on host computers. The data can be exchanged via email and chat.

Question 6.
What is an Interactive Operating system?
Answer:
This is the operating system that provides a GUI through which the user can navigate and interact.

Question 7.
Explain Round Robin Scheduling.
Answer:
This type of scheduling is also known as Time sharing scheduling process. In this, each program is given a fixed amount of time to execute.

Question 8.
What is Memory Management?
Answer:
It is the main functionality of an operating system which handles or manages primary memory and moves processes back and forth between main memory and disk during execution.

Question 9.
Mention different management techniques?
Answer:
Single continuous allocation, Partitioned allocation, Paged memory management, Segmented memory management.

Question 10.
What is Linux?
Answer:
Linux is a family of open-source operating systems.

Question 11.
What is an Android?
Answer:
Android is a mobile operating system developed by Google, based on the Linux and designed primarily for touch screen mobile devices such as smart phones and tablets.

Question 12.
What are the types of software?
Answer:
Application software and System software

Question 13.
What is Process Management?
Answer:
Process management is a function that includes creating and deleting processes and providing mechanisms for processes to communicate and synchronize with each other.

Question 14.
What is Round Robin Scheduling?
Answer:
This algorithm is designed especially for time sharing systems.

Question 15.
What are the 3 levels of security?
Answer:

  1. File Access Level
  2. System Level
  3. Network Level.

Question 16.
Name an OS which is Multitasking and Multi – user operating system.
Answer:
UNIX.

PART – 3
III. Explain in Brief

Question 1.
Write a short note on Android.
Answer:
Android:
Android is a mobile operating system developed by Google, based on Linux and designed primarily for touch screen mobile devices such as smart phones and tablets. Google has further developed Android TV for televisions, Android Auto for cars and Android Wear for wrist watches, each with a specialized user interface. Variants of Android are also used on game consoles, digital cameras, PCs and other electronic gadgets.

Question 2.
Why the operating system is needed?
Answer:
Need for operating system:
Operating System has become essential to enable the users to design applications without the knowledge of the computer’s internal structure of hardware. Operating System manages all the Software and Hardware. Most of the time there are many different computer programmes running at the same time, they all need to access the Computers, CPU, Memory and Storage. The need of Operating System is basically – an interface between the user and hardware.
Computer Operating System In Tamil Pdf Samacheer Kalvi 11th Computer Science Solutions Chapter 4
Operating System works as translator, while it translates the user request into machine language(Binary language), processes it and then sends it back to Operating System. Operating System converts processed information into user readable form.

Question 3.
Write a note on OS for mobile devices.
Answer:
Operating systems for mobile devices:
Mobile devices such as phones, tablets and MP3 players are different from desktop and laptop computers and hence they need special Operating Systems. Examples of mobile Operating Systems are Apple iOS and Google Android. The iOS running on an iPad is Operating systems for mobile devices generally are not as fully featured as those made for desktop and laptop computers and they are not able to run all software.

Question 4.
Write a note on iOS – iphone OS.
Answer:
iOS (formerly iPhone OS) is a mobile Operating System created and developed by Apple Inc., exclusively for its hardware. It is the Operating System that presently powers many of the company’s mobile devices, including the iPhone, iPad and iPod Touch. It is the second most popular mobile Operating System globally after Android.

Question 5.
What is a software? Explain its types in detail.
Answer:
A software is a set of instructions that perform specific task. It interacts basically with the hardware to generate the desired output.
Types of Software:
Software is classified into two types:

  1. Application Software
  2. System Software

Application Software:
Application software is a set of programs to perform specific task. For example MS-word is an application software to create text document and VLC player is familiar application software to play audio, video files and many more.

System Software:
System software is a type of computer program that is designed to run the computer’s hardware and application programs. For example Operating System and Language Processor.

Question 6.
Write note on single user OS?
Answer:
An os allows only a single user to perform a task at a time. It is called as a single user and single task os.
Example : MS – DOS.

Question 7.
Write note on Raspbion os?
Answer:
Raspbion os is a platform that is designed to teach how to build a computer, working principle of every part of a circuit board, write code apps or games. The platform is available in pre – designed kits.

Question 8.
Explain the memory management activities done by os?
Answer:

  1. Keeping track of which portion of memory are currently being used and who is using them.
  2. Determining which processes and data to more in and out of memory.
  3. Allocation and de – allocation of memory blocks as needed by the program in main memory. (Garbage collection).

Question 9.
Define Process?
Answer:

  1. A process is the unit of work (program) in a computer.
  2. A word processing program being run by an individual user on a computer is a process.
  3. A system task, such as sending output to printer or screen can also be called as a process.

Question 10.
How are the processes classified on process management?
Answer:
A computer consists of a collection of process they are classified as two categories:

  1. Operating system process which is executed by system code.
  2. User processes which is executed by user code.

Question 11.
Name the activities done by os related with the process management?
Answer:

  1. Scheduling processes and threads on the cpu.
  2. Creating and deleting both user and system processes.
  3. Suspending and resuming processes.
  4. Providing mechanisms for process synchronization.
  5. Providing mechanisms for process communication.

Question 12.
Write note an Fault Tolerance?
Answer:

  1. The operating systems should be robust.
  2. When there is a fault, the operating system should not crash, instead the os have fault tolerance capabilities and retain the existing state of system.

Question 13.
Write note on File Allocations Table (FAT).
Answer:

  1. Any type of data in a computer is stored in the form of flies and directories / folders through File Allocation Table (FAT).
  2. The FAT stores general information about files like file name, type (text or binary), size, starting address and access mode (sequential / indexed / indexed – sequential / direct / relative).

Question 14.
Give an example for Time – Sharing concept?
Answer:
Let us assume that there are three processes called P1, P2, P3 and time allocated for each process 30,40, 50 minutes. If the process P1 completes with in 20 minutes then processor takes the next Process P1 for the execution. If the process P2 could not complete within 40 minutes, then the current process P2 will be paused and switch over to the next process P3.

Question 15.
Write note on React OS.
React os is a windows – alternative open source os which is being developed on the principles of windows – without using any of the microsoft code.

Question 16.
List any 6 Android Mobile Open Source OS.
Answer:

  1. Honey comb
  2. Jelly Bean
  3. Kitkat
  4. Lollipop
  5. Marshmallow
  6. Nougat

PART – 4
IV. Explain in Detail

Question 1.
Write six uses of operating system.
Answer:
The main use of Operating System is

  1. to ensure that a computer can be used to extract what the user wants it do.
  2. Easy interaction between the users and computers.
  3. Starting computer operation automatically when power is turned on (Booting).
  4. Controlling Input and Output Devices
  5. Manage the utilization of main memory.
  6. Providing security to user programs.

Question 2.
Draw the diagram for the key features of the operating system.
Answer:
Samacheer Kalvi Advantages And Disadvantages 11th Computer Science Solutions Chapter 4 Theoretical Concepts Of Operating System

Question 3.
Explain User Interface?
Answer:
User Interface:
User interface is one of the significant feature in Operating System. The only way that user can make interaction with a computer. If the computer interface is not user – friendly, the user slowly reduces the computer usage from their normal life. This is the main reason for the key success of GUI (Graphical User Interface) based Operating System. The GUI is a window based system with a pointing device to direct I/O, choose from menus, make selections and a keyboard to enter text. Its vibrant colours attract the user very easily. Beginners are impressed by the help and pop up window message boxes. Icons are playing vital role of the particular application.

Now Linux distribution is also available as GUI based Operating System. The following points are considered when User Interface is designed for an application.

  1. The user interface should enable the user to retain this expertise for a longer time.
  2. The user interface should also satisfy the customer based on their needs.
  3. The user interface should save user’s precious time. Create graphical elements like Menus,Window,Tabs, Icons and reduce typing work will be an added advantage of the Operating System.
  4. The ultimate aim of any product is to satisfy the customer. The User Interface is also designed to satisfy the customer.
  5. The user interface should reduce number of errors committed by the user with a little practice the user should be in a position to avoid errors (Error Log File)

Question 4.
Explain Memory Management?
Answer:
Memory Management:
Memory Management is the process of controlling and coordinating computer’s main memory and assigning memory block (space) to various running programs to optimize overall computer performance. The Memory management involves the allocation of specific memory blocks to individual programs based on user demands. At the application level, memory management ensures the availability of adequate memory for each running program at all times.

The objective of Memory Management process is to improve both the utilization of the CPU and the speed of the computer’s response to its users via main memory. For these reasons the computers must keep several programs in main memory that associates with many different Memory Management schemes.

The Operating System is responsible for the following activities in connection with memory management:

  1. Keeping track of which portion of memory are currently being used and who is using them.
  2. Determining which processes (or parts of processes) and data to move in and out of memory.
  3. Allocation and de – allocation of memory blocks as needed by the program in main memory. (Garbage Collection)

Question 5.
Explain process Management.
Answer:
Process Management:
Process management is function that includes creating and deleting processes and providing mechanisms for processes to communicate and synchronize with each other. A process is the unit of work (program) in a computer. A word processing program being run by an individual user on a computer is a process. A system task, such as sending output to a printer or screen, can also be called as a Process.

A computer consists of a collection of processes, they are classified as two categories:

  1. Operating System processes which is executed by system code.
  2. User Processes which is execute by user code.

All these processes can potentially execute concurrently on a single CPU. A process needs certain resources including CPU time, memory, files and I/O devices to finish its task.

The Operating System is responsible for the following activities associated with the process management:

  1. Scheduling processes and threads on the CPUs.
  2. Creating and deleting both user and system processes.
  3. Suspending and resuming processes.
  4. Providing mechanisms for process synchronization.
  5. Providing mechanisms for process communication.

The following algorithms are mainly used to allocate the job (process) to the processor.

  1. FIFO
  2. SJF
  3. Round Robin
  4. Based on Priority

Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

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Samacheer Kalvi 11th Physics Chapter 3 Laws of Motion Textual Questions Solved

Samacheer Kalvi 11th Physics Laws of Motion Multiple Choice Questions

11th Physics Chapter 3 Book Back Answers Question 1.
When a car takes a sudden left turn in the curved road, passengers are pushed towards the right due to
(a) inertia of direction
(b) inertia of motion
(c) inertia of rest
(d) absence of inertia
Answer:
(a) inertia of direction

Samacheer Kalvi 11th Physics Solution Chapter 3 Question 2.
An object of mass m held against a vertical wall by applying horizontal force F as shown in the figure. The minimum value of the force F is [IIT JEE 1994]
(a) Less than mg
(b) Equal to mg
(c) Greater than mg
(d) Cannot determine
Answer:
(c) Greater than mg

11th Physics Lesson 3 Book Back Answers Question 3.
A vehicle is moving along the positive x direction, if sudden brake is applied, then
(a) frictional force acting on the vehicle is along negative x direction
(b) frictional force acting on the vehicle is along positive x direction
(c) no frictional force acts on the vehicle
(d) frictional force acts in downward direction
Answer:
(a) frictional force acting on the vehicle is along negative x direction

11th Physics 3rd Chapter Book Back Answers Question 4.
A book is at rest on the table which exerts a normal force on the book. If this force is considered as reaction force, what is the action force according to Newton’s third law?
(a) Gravitational force exerted by Earth on the book
(b) Gravitational force exerted by the book on Earth
(c) Normal force exerted by the book on the table
(d) None of the above
Answer:
(c) Normal force exerted by the book on the table

Laws Of Motion Class 11 State Board Question 5.
Two masses m1 and m2 are experiencing the same force where m1 < m2 The ratio of their acceleration \(\frac{a_{1}}{a_{2}}\) is –
(a) 1
(b) less than 1
(c) greater than 1
(d) all the three cases
Answer:
(c) greater than 1

Samacheer Kalvi Guru 11th Physics Question 6.
Choose appropriate free body diagram for the particle experiencing net acceleration along negative y direction. (Each arrow mark represents the force acting on the system).
11th Physics Chapter 3 Book Back Answers Samacheer Kalvi Laws Of Motion
Answer:
Samacheer Kalvi 11th Physics Solution Chapter 3 Laws Of Motion

Class 11 Physics Solutions Samacheer Kalvi Question 7.
A particle of mass m sliding on the smooth double inclined plane (shown in figure) will experience –
(a) greater acceleration along the path AB
(b) greater acceleration along the path AC
(c) same acceleration in both the paths
(d) no acceleration in both the paths
11th Physics Lesson 3 Book Back Answers Samacheer Kalvi Laws Of Motion
Answer:
(a) greater acceleration along the path AC

Samacheer Kalvi 11th Physics Question 8.
Two blocks of masses m and 2m are placed on a smooth horizontal surface as shown. In the first case only a force F1  is applied from the left. Later only a force F2  is applied from the right. If the force acting at the interface of the two blocks in the two cases is same, then F1 : F2 is [Physics Olympiad 2016]
11th Physics 3rd Chapter Book Back Answers Samacheer Kalvi Laws Of Motion
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 1 : 3
Answer:
(c) 2 : 1

11th Physics 3rd Lesson Book Back Answers Question 9.
Force acting on the particle moving with constant speed is –
(a) always zero
(b) need not be zero
(c) always non zero
(d) cannot be concluded
Answer:
(b) need not be zero

11th Physics 3rd Chapter Exercise Question 10.
An object of mass m begins to move on the plane inclined at an angle 0. The coefficient of static friction of inclined surface is lay. The maximum static friction experienced by the mass is –
(a) mg
(b) µs mg
(c) µs mg sin θ
(d) µs mg cos θ
Answer:
(d) µs mg cos θ

Class 11 Samacheer Physics Solutions Question 11.
When the object is moving at constant velocity on the rough surface –
(a) net force on the object is zero
(b) no force acts on the object
(c) only external force acts on the object
(d) only kinetic friction acts on the object
Answer:
(a) net force on the object is zero

Samacheer Kalvi 11th Physics Book Back Answers Question 12.
When an object is at rest on the inclined rough surface –
(a) static and kinetic frictions acting on the object is zero
(b) static friction is zero but kinetic friction is not zero
(c) static friction is not zero and kinetic friction is zero
(d) static and kinetic frictions are not zero
Answer:
(c) static friction is not zero and kinetic friction is zero

Class 11 Physics Samacheer Kalvi Question 13.
The centrifugal force appears to exist –
(a) only in inertial frames
(b) only in rotating frames
(c) in any accelerated frame
(d) both in inertial and non-inertial frames
Answer:
(b) only in rotating frames

11th Physics Unit 3 Book Back Answers Question 14.
Choose the correct statement from the following –
(a) Centrifugal and centripetal forces are action reaction pairs
(b) Centripetal forces is a natural force
(c) Centrifugal force arises from gravitational force
(d) Centripetal force acts towards the center and centrifugal force appears to act away from the center in a circular motion
Answer:
(d) Centripetal force acts towards the center and centrifugal force appears to act away from the center in a circular motion

Laws Of Motion Class 11 Numericals With Solutions Pdf Question 15.
If a person moving from pole to equator, the centrifugal force acting on him –
(a) increases
(b) decreases
(c) remains the same
(d) increases and then decreases
Answer:
(a) increases

Samacheer Kalvi 11th Physics Laws of Motion Short Answer Questions

Question 1.
Explain the concept of inertia. Write two examples each for inertia of motion, inertia of rest and inertia of direction.
Answer:
The inability of objects to move on its own or change its state of motion is called inertia. Inertia means resistance to change its state. There are three types of inertia:
1. Inertia of rest:
The inability of an object to change its state of rest is called inertia of rest.
Example:

  • When a stationary bus starts to move, the passengers experience a sudden backward push.
  • A book lying on the table will remain at rest until it is moved by some external agencies.

2. Inertia of motion:
The inability of an object to change its state of uniform speed (constant speed) on its own is called inertia of motion.
Example:

  • When the bus is in motion, and if the brake is applied suddenly, passengers move forward and hit against the front seat.
  • An athlete running is a race will continue to run even after reaching the finishing point.

3. Inertia of direction:
The inability of an object to change its direction of motion on its own is called inertia of direction.
Example:

  • When a stone attached to a string is in whirling motion, and if the string is cut suddenly, the stone will not continue to move in circular motion but moves tangential to the circle.
  • When a bus moving along a straight line takes a turn to the right. The passengers are thrown towards left.

Question 2.
State Newton’s second law.
Answer:
The force acting on an object is equal to the rate of change of its momentum –
\(\overline{\mathrm{F}}\) = \(\frac{d \bar{p}}{d t}\)

Question 3.
Define one newton.
Answer:
One newton is defined as the force which acts on 1 kg of mass to give an acceleration 1 ms-2 in the direction of the force.

Question 4.
Show that impulse is the change of momentum.
Answer:
According to Newton’s Second Law
F = \(\frac {dp}{dt}\) i.e. dp = Fdt
Integrate it over a time interval from ti  to tf
Laws Of Motion Class 11 State Board Samacheer Kalvi Chapter 3 Laws Of Motion
Pi → initial momentum of the object at ti
Pf → Final momentum of the object at tf
Pf – Pi = ∆p = change in momentum during the time interval ∆t.
\(\int_{t_{i}}^{t_{f}} \mathrm{F} \cdot d t=\mathrm{J}\) is called the impulse.
If the force is constant over the time interval ∆t, then
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Hence the proof.

Question 5.
Using free body diagram, show that it is easy to pull an object than to push it.
Answer:
When a body is pushed at an arbitrary angle θ [0 to \(\frac {π}{2}\)], the applied force F can be resolved into two components as F sin 0 parallel to the surface and F cos 0 perpendicular to the surface as shown in figure. The total downward force acting on the body is mg + F cos θ. It implies that the normal force acting on the body increases. Since there is no acceleration along the vertical direction the normal force N is equal to
Npush = mg + F cos θ …………(1)
As a result the maximal static friction also increases and is equal to
\(f_{S}^{\max }\) = \(\mu_{r} \mathrm{N}_{\mathrm{push}}\) = µs(mg + F cos θ) ……(2)
Equation (2) shows that a greater force needs to be applied to push the object into motion.
Samacheer Kalvi Guru 11th Physics Solutions Chapter 3 Laws Of Motion

When an object is pulled at an angle θ, the applied force is resolved into two components as shown in figure. The total downward force acting on the object is –
Npull = mg – F cos θ ………….(3)
Class 11 Physics Solutions Samacheer Kalvi Chapter 3 Laws Of Motion

Equation (3) shows that the normal force is less than – Npush. From equations (1) and (3), it is easier to pull an object than to push to make it move.

Question 6.
Explain various types of friction. Suggest a few methods to reduce friction.
Answer:
There are two types of Friction:
(1) Static Friction:
Static friction is the force which opposes the initiation of motion of an object on the surface. The magnitude of static frictional force fs lies between
\(0 \leq f_{s} \leq \mu_{s} \mathrm{N}\)
where, µs – coefficient of static friction
N – Normal force

(2) Kinetic friction:
The frictional force exerted by the surface when an object slides is called as kinetic friction. Also called as sliding friction or dynamic friction,
fk – µkN
where µk – the coefficient of kinetic friction
N – Normal force exerted by the surface on the object

Methods to reduce friction:
Friction can be reduced

  • By using lubricants
  • By using Ball bearings
  • By polishing
  • By streamlining

Question 7.
What is the meaning by ‘pseudo force’?
Answer:
Pseudo force is an apparent force which has no origin. It arises due to the non-inertial nature of the frame considered.

Question 8.
State the empirical laws of static and kinetic friction.
Answer:
The empirical laws of friction are:

  • Friction is independent of surface of contact.
  • Coefficient of kinetic friction is less than coefficient of static friction.
  • The direction of frictional force is always opposite to the motion of one body over the other.
  • Frictional force always acts on the object parallel to the surface on which the objet is placed,
  • The magnitude of frictional force between any two bodies in contact is directly proportional to the normal reaction between them.

Question 9.
State Newton’s third law.
Answer:
Newton’s third law states that for every action there is an equal and opposite reaction.

Question 10.
What are inertial frames?
Answer:
Inertial frame is the one in which if there is no force on the object, the object moves at constant velocity.

Question 11.
Under what condition will a car skid on a leveled circular road?
Answer:
On a leveled circular road, if the static friction is not able to provide enough centripetal ’force to turn, the vehicle will start to skid
\(\mu_{s}<\frac{v^{2}}{r g}\)

Samacheer Kalvi 11th Physics Laws of Motion Long Answer Questions

Question 1.
Prove the law of conservation of linear momentum. Use it to find the recoil velocity of a gun when a bullet is fired from it.
Answer:
In nature, conservation laws play a very important role. The dynamics of motion of bodies can be analysed very effectively using conservation laws. There are three conservation laws in mechanics. Conservation of total energy, conservation of total linear momentum, and conservation of angular momentum. By combining Newton’s second and third laws, we can derive the law of conservation of total linear momentum. When two particles interact with each other, they exert equal and opposite forces on each other.

The particle 1 exerts force \(\overrightarrow{\mathrm{F}}_{12}\) on particle 2 and particle 2 exerts an exactly equal and opposite force \(\overrightarrow{\mathrm{F}}_{12}\) on particle 1 according to Newton’s third law.
\(\overrightarrow{\mathrm{F}}_{12}\) = –\(\overrightarrow{\mathrm{F}}_{12}\) ……..(1)
In terms of momentum of particles, the force on each particle (Newton’s second law) can be written as –
\(\overrightarrow{\mathrm{F}}_{12}\) = \(\frac{d \vec{p}_{1}}{d t}\) and \(\overrightarrow{\mathrm{F}}_{21}\) = \(\frac{d \vec{p}_{2}}{d t}\) ………(2)
Here \(\vec{p}_{1}\) is the momentum of particle 1 which changes due to the force \(\overrightarrow{\mathrm{F}}_{12}\) exerted by particle 2. Further Po is the momentum of particle \(\vec{p}_{2}\) This changes due to \(\overrightarrow{\mathrm{F}}_{21}\) exerted by particle 1.
Substitute equation (2) in equation (1)
\(\frac{d \vec{p}_{1}}{d t}\) = – \(\frac{d \vec{p}_{2}}{d t}\) …………(3)
\(\frac{d \vec{p}_{1}}{d t}\) + \(\frac{d \vec{p}_{2}}{d t}\) = 0 ………(4)
\(\frac {d}{dt}\)(\(\vec{p}_{1}\) + \(\vec{p}_{2}\)) = 0
It implies that \(\vec{p}_{1}\) + \(\vec{p}_{2}\) = constant vector (always).
\(\vec{p}_{1}\) + \(\vec{p}_{2}\) is the total linear momentum of the two particles (\(\vec{p}_{tot}\) = \(\vec{p}_{1}\) + \(\vec{p}_{2}\)).It is also called as total linear momentum of the system. Here, the two particles constitute the system. From this result, the law of conservation of linear momentum can be stated as follows.

If there are no external forces acting on the system, then the total linear momentum of the system (\(\vec{p}_{tot}\)) is always a constant vector. In other words, the total linear momentum of the system is conserved in time. Here the word ‘conserve’ means that \(\vec{p}_{1}\) and \(\vec{p}_{2}\) can vary, in such a way that \(\vec{p}_{1}\) + \(\vec{p}_{2}\) is a constant vector.

The forces \(\overrightarrow{\mathrm{F}}_{12}\) and \(\overrightarrow{\mathrm{F}}_{21}\) are called the internal forces of the system, because they act only between the two particles. There is no external force acting on the two particles from outside. In such a case the total linear momentum of the system is a constant vector or is conserved.

Meaning of law of conservation of momentum:
1. The Law of conservation of linear momentum is a vector law. It implies that both the magnitude and direction of total linear momentum are constant. In some cases, this total momentum can also be zero.

2. To analyse the motion of a particle, we can either use Newton’s second law or the law of conservation of linear momentum. Newton’s second law requires us to specify the forces involved in the process. This is difficult to specify in real situations. But conservation of linear momentum does not require any force involved in the process. It is convenient and hence important.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws Of Motion

For example, when two particles collide, the forces exerted by these two particles on each other is difficult to specify. But it is easier to apply conservation of linear momentum during the collision process.

Examples:
Consider the firing of a gun. Here the system is Gun+bullet. Initially the gun and bullet are at rest, hence the total linear momentum of the system is zero. Let \(\vec{p}_{1}\) be the momentum of the bullet and \(\vec{p}_{2}\) the momentum of the gun before firing. Since initially both are at rest,

11th Physics 3rd Lesson Book Back Answers Laws Of Motion Samacheer Kalvi
Total momentum before firing the gun is zero, \(\vec{p}_{1}\) + \(\vec{p}_{2}\) = 0.
According to the law of conservation of linear momentum, total linear momentum has to be zero after the firing also.

When the gun is fired, a force is exerted by the gun on the bullet in forward direction. Now the momentum of the bullet changes from \(\vec{p}_{1}\) to \(\vec{p}_{1}\) To conserve the total linear momentum of the system, the momentum of the gun must also change from \(\vec{p}_{2}\) to \(\vec{p}_{2}^{\prime}\). Due to the conservation of linear momentum, \(\vec{p}_{1}\) + \(\vec{p}_{2}^{\prime}\) = 0.

It implies that \(\vec{p}_{1}^{\prime}\)= \(\vec{p}_{2}^{\prime}\), the momentum of the gun is exactly equal, but in the opposite direction to the momentum of the bullet. This is the reason after firing, the gun suddenly moves backward with the momentum (-\(\vec{p}_{2}^{\prime}\)). It is called ‘recoil momentum’. Th is is an example of conservation of total linear momentum.
11th Physics 3rd Chapter Exercise Samacheer Kalvi Laws Of Motion

Question 2.
What are concurrent forces? State Lami’s theorem.
Answer:
Concurrent force:
A collection of forces is said to be concurrent, if the lines of forces act at a common point.

Lami’s Theorem:
If a system of three concurrent and coplanar forces is in equilibrium, then Lami’s theorem states that the magnitude of each force of the system is proportional to sine of the angle between the other two forces.
i.e. |\(\overrightarrow{\mathrm{F}}_{1}\)|∝ sin α, |\(\overrightarrow{\mathrm{F}}_{2}\)| ∝ sin β, |\(\overrightarrow{\mathrm{F}}_{2}\)| ∝ sin γ,

Question 3.
Explain the motion of blocks connected by a string in

  1. Vertical motion
  2. Horizontal motion.

Answer:
When objects are connected by strings and When objects are connected by strings and a force F is applied either vertically or horizontally or along an inclined plane, it produces a tension T in the string, which affects the acceleration to an extent. Let us discuss various cases for the same.

Case 1:
Vertical motion:
Consider two blocks of masses m1 and m2 (m1> m2) connected by a light and in-extensible string that passes over a pulley as shown in Figure.

Class 11 Samacheer Physics Solutions Chapter 3 Laws Of Motion

Let the tension in the string be T and acceleration a. When the system is released, both the blocks start moving, m2 vertically upward and mk, downward with same acceleration a. The gravitational force m1g on mass m1 is used in lifting the mass m2. The upward direction is chosen as y direction. The free body diagrams of both masses are shown in Figure.
Samacheer Kalvi 11th Physics Book Back Answers Chapter 3 Laws Of Motion

Applying Newton’s second law for mass m2 T \(\hat{j}\) – m2 g \(\hat{j}\) = m2 a \(\hat{j}\) The left hand side of the above equation is the total force that acts on m2 and the right hand side is the product of mass and acceleration of m2 in y direction.
By comparing the components on both sides, we get
T = m2 g = m2 a ……….(1)
Similarly, applying Newton’s second law for mass m2
T \(\hat{j}\) – m1 g\(\hat{j}\) = m1a\(\hat{j}\)
As mass m1 moves downward (-\(\hat{j}\)), its acceleration is along (-\(\hat{j}\))
By comparing the components on both sides, we get
T = m1g = -m1a
m1g – T = m1a ………..(2)
Adding equations (1) and (2), we get
m1g – m2g = m1a + m2a
(m1 – m2)g = (m1 + m2)a …………(3)
From equation (3), the acceleration of both the masses is –
a = (\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))g ………..(4)
If both the masses are equal (m1 = m2), from equation (4)
a = 0
This shows that if the masses are equal, there is no acceleration and the system as a whole will be at rest.
To find the tension acting on the string, substitute the acceleration from the equation (4) into the equation (1).
T = m2g = m2(\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))
T = m2g + m2 (\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))g ……….(5)
By taking m2g common in the RHS of equation (5)
Class 11 Physics Samacheer Kalvi Solutions Chapter 3 Laws Of Motion
Equation (4) gives only magnitude of acceleration.
For mass m1, the acceleration vector is given by \(\vec{a}\) = –\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\)\(\hat{j}\)
For mass m2, the acceleration vector is given by \(\vec{a}\) = \(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\) \(\hat{j}\)

Case 2:
Horizontal motion:
In this case, mass m2 is kept on a horizontal table and mass m1, is hanging through a small pulley as shown in figure. Assume that there is no friction on the surface
11th Physics Unit 3 Book Back Answers Samacheer Kalvi Laws Of Motion
As both the blocks are connected to the un stretchable string, if m1 moves with an acceleration a downward then m2 also moves with the same acceleration a horizontally.
The forces acting on mass m2 are

  • Downward gravitational force (m2g)
  • Upward normal force (N) exerted by the surface
  • Horizontal tension (T) exerted by the string

The forces acting on mass m1 are

  • Downward gravitational force (m1g)
  • Tension (T) acting upwards

The free body diagrams for both the masses is shown in figure.

Laws Of Motion Class 11 Numericals With Solutions Pdf Samacheer Kalvi
Applying Newton’s second law for m1
T\(\hat{i}\) – m1g\(\hat{j}\) = -m1a\(\hat{j}\) (alongy direction)
By comparing the components on both sides of the above equation,
T – m1g = -m1a …………(1)
Applying Newton’s second law for m2
Ti = m1ai (along x direction)
By comparing the components on both sides of above equation,
T = m2a ………….(2)
There is no acceleration along y direction for m2.
N\(\hat{j}\) – m2g\(\hat{j}\) = 0
By comparing the components on both sides of the above equation
N – m2g = 0
N = m2g ……….(3)
By substituting equation (2) in equation (1), we can find the tension T
m2a – m1g = -m1a
m2a + m1a = m1g
a = \(\frac{m_{1}}{m_{1}+m_{2}}\)g …………(4)
Tension in the string can be obtained by substituting equation (4) in equation (2)
T = \(\frac{m_{1} m_{2}}{m_{1}+m_{2}}\)g ………..(5)

Comparing motion in both cases, it is clear that the tension in the string for horizontal motion is half of the tension for vertical motion for same set of masses and strings. This result has an important application in industries. The ropes used in conveyor belts (horizontal motion) work for longer duration than those of cranes and lifts (vertical motion).

Question 4.
Briefly explain the origin of friction. Show that in an inclined plane, angle of friction is pqual to angle of repose.
Answer:
If a very gentle force in the horizontal direction is given to an object at rest on the table it does not move. It is because of the opposing force exerted by the surface on the object which resists its motion. This force is called the frictional force. During the time of Newton and Galileo, frictional force was considered as one of the natural forces like gravitational force. But in the twentieth century, the understanding on atoms, electron and protons has changed the perspective.

The frictional force is actually the electromagnetic force between the atoms on the two surfaces. Even well polished surfaces have irregularities on the surface at the microscopic level. The component of force parallel to the inclined plane (mg sin θ) tries to move the object down. The component of force perpendicular to the inclined plane (mg cos θ) is balanced by the Normal force (N).
N = mg cos θ ………(1)

When the object just begins to move, the static friction attains its maximum value
fs = \(f_{s}^{\max }\) = µs N
This friction also satisfies the relation
\(f_{s}^{\max }\) = µs mg sin θ ……….(2)
Equating the right hand side of equations (1) and (2),
(\(f_{s}^{\max }\))/N = sin θ / cos θ
From the definition of angle of friction, we also know that
tan θ = µs ………..(3)
in which θ is the angle of friction.
Thus the angle of repose is the same as angle of friction. But the difference is that the angle of repose refers to inclined surfaces and the angle of friction is applicable to any type of surface.

Question 5.
State Newton’s three laws and discuss their significance.
Answer:
First Law:
Every object continues to be in the state of rest or of uniform motion (constant velocity) unless there is external force acting on it.

Second Law:
The force acting on an object is equal to the rate of change of its momentum

Third Law:
For every action there is an equal and opposite reaction.

Significance of Newton’s Laws:
1. Newton’s laws are vector laws. The equation \(\overline{\mathrm{F}}\) = m\(\overline{\mathrm{a}}\) is a vector equation and essentially it is equal to three scalar equations. In Cartesian coordinates, this equation can be written as Fx\(\hat{i}\) + Fy\(\hat{j}\) + Fz\(\hat{j}\) = max\(\hat{i}\) + may\(\hat{j}\) + maz\(\hat{j}\)
By comparing both sides, the three scalar equations are

Fx = max The acceleration along the x-direction depends only on the component of force acting along the x – direction.
Fy = may The acceleration along the y direction depends only on the component of force acting along the y – direction.
Fz = maz The acceleration along the z direction depends only on the component of force acting along the z – direction.
From the above equations, we can infer that the force acting along y direction cannot alter the acceleration along x direction. In the same way, Fz cannot affect ay and ax. This understanding is essential for solving problems.

2. The acceleration experienced by the body at time t depends on the force which acts on the body at that instant of time. It does not depend on the force which acted on the body before the time t. This can be expressed as
\(\overline{\mathrm{F}}\)(t) = m\(\overline{\mathrm{a}}\)(t)
Acceleration of the object does not depend on the previous history of the force. For example, when a spin bowler or a fast bowler throws the ball to the batsman, once the ball leaves the hand of the bowler, it experiences only gravitational force and air frictional force. The acceleration of the ball is independent of how the ball was bowled (with a lower or a higher speed).

3. In general, the direction of a force may be different from the direction of motion. Though in some cases, the object may move in the same direction as the direction of the force, it is not always true. A few examples are given below.

Case 1:
Force and motion in the same direction:
When an apple falls towards the Earth, the direction of motion (direction of velocity) of the apple and that of force are in the same downward direction as shown in the Figure.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Case 2:
Force and motion not in the same direction:
The Moon experiences a force towards the Earth. But it actually moves in elliptical orbit. In this case, the direction of the force is different from the direction of motion as shown in Figure.

Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Case 3:
Force and motion in opposite direction:
If an object is thrown vertically upward, the direction of motion is upward, but gravitational force is downward as shown in the Figure.

Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Case 4:
Zero net force, but there is motion:
When a raindrop gets detached from the cloud it experiences both downward gravitational force and upward air drag force. As it descends towards the Earth, the upward air drag force increases and after a certain time, the upward air drag force cancels the downward gravity. From then on the raindrop moves at constant velocity till it touches the surface of the Earth. Hence the raindrop comes with zero net force, therefore with zero acceleration but with non-zero terminal velocity. It is shown in the Figure

Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

4. If multiple forces \(\overrightarrow{\mathrm{F}}_{1}\), \(\overrightarrow{\mathrm{F}}_{2}\), \(\overrightarrow{\mathrm{F}}_{3}\),……. \(\overrightarrow{\mathrm{F}}_{n}\) act on the same body, then the total force (\(\overrightarrow{\mathrm{F}}_{net}\)) is equivalent to the vectorial sum of the individual forces. Their net force provides the acceleration.
\(\overrightarrow{\mathrm{F}}_{net}\) = \(\overrightarrow{\mathrm{F}}_{1}\) + \(\overrightarrow{\mathrm{F}}_{2}\) + \(\overrightarrow{\mathrm{F}}_{3}\) + ……… + \(\overrightarrow{\mathrm{F}}_{n}\)
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Newton’s second law for this case is –
\(\overrightarrow{\mathrm{F}}_{net}\) = m\(\overline{\mathrm{a}}\)
In this case the direction of acceleration is in the direction of net force.
Example:
Bow and arrow

Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

5. Newton’s second law can also be written in the following form.
Since the acceleration is the second derivative of position vector of the body \(\left(\vec{a}=\frac{d^{2} \vec{r}}{d t^{2}}\right)\)
the force on the body is –
\(\overline{\mathrm{F}}\) = m\(\frac{d^{2} \vec{r}}{d t^{2}}\)
From this expression, we can infer that Newton’s second law is basically a second order ordinary differential equation and whenever the second derivative of position vector is not zero, there must be a force acting on the body.

6. If no force acts on the body then Newton’s second law, m = \(\frac{d \vec{v}}{d t}\) = 0
It implies that \(\overline{\mathrm{v}}\) = constant. It is essentially Newton’s first law. It implies that the second law is consistent with the first law. However, it should not be thought of as the reduction of second law to the first when no force acts on the object. Newton’s first and second laws are independent laws. They can internally be consistent with each other but cannot be derived from each other.

7. Newton’s second law is cause and effect relation. Force is the cause and acceleration is the effect. Conventionally, the effect should be written on the left and cause on the right hand side of the equation. So the correct way of writing Newton’s second law is –
m\(\overline{\mathrm{a}}\) = \(\overline{\mathrm{F}}\) or \(\frac{d \vec{p}}{d t}\) = \(\overline{\mathrm{F}}\)

Question 6.
Explain the similarities and differences of centripetal and centrifugal forces.
Answer:
Salient features of centripetal and centrifugal forces.
Centripetal Force:

  • It is a real force which is exerted on the body by the external agencies like gravitational force, tension in the string, normal force etc.
  • Acts in both inertial and non-inertial frames
  • It acts towards the axis of rotation or center of the circle in circular motion
    \(\left|\overrightarrow{\mathrm{F}}_{\mathrm{C}_{\mathrm{P}}}\right|\) = mω2r = \(\frac{m v^{2}}{r}\)
  • Real force and has real effects
  • Origin of centripetal force is interaction between two objects.
  • In inertial frames centripetal force has to be included when free body diagrams are drawn.

Centrifugal Force:

  • It is a pseudo force or fictitious force which cannot arise from gravitational force, tension force, normal force etc.
  • Acts only in rotating frames (non-inertial frame)
  • It acts outwards from the axis of rotation or radially outwards from the center of the circular motion
    \(\left|\overrightarrow{\mathrm{I}}_{\mathrm{C}_{\mathrm{f}}}\right|\) = mω2r = \(\frac{m v^{2}}{r}\)
  • Pseudo force but has real effects
  • Origin of centrifugal force is inertia. It does not arise from interaction. In an inertial frame the object’s inertial motion appears as centrifugal force in the rotating frame.
  • In inertial frames there is no centrifugal force. In rotating frames, both centripetal and centrifugal force have to be included when free body diagrams are drawn.

Question 7.
Briefly explain ‘centrifugal force’ with suitable examples.
Answer:
To use Newton’s first and second laws in the rotational frame of reference, we need to include a Pseudo force called centrifugal force. This centrifugal force appears to act on the object with respect to rotating frames.

Circular motion can be analysed from two different frames of reference. One is the inertial frame (which is either at rest or in uniform motion) where Newton’s laws are obeyed. The other is the rotating frame of reference which is a non – inertial frame of reference as it is accelerating.

When we examine the circular motion from these frames of reference the situations are entirely different. To use Newton’s first and second laws in the rotational frame of reference, we need to include a pseudo force called ‘centrifugal force’. This ‘centrifugal force’ appears to act on the object with respect to rotating frames. To understand the concept of centrifugal force, we can take a specific case and discuss as done below.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Free body diagram of a particle including the centrifugal force Consider the case of a whirling motion of a stone tied to a string. Assume that the stone has angular velocity ω in the inertial frame (at rest). If the motion of the stone is observed from a frame which is also rotating along with the stone with same angular velocity ω then, the stone appears to be at rest.

This implies that in addition to the inward centripetal force – mω2r there must be an equal and opposite force that acts on the stone outward with value +mω2r . So the total force acting on the stone in a rotating frame is equal to zero (-mω2r + mω2 r = 0). This outward force +mω2r is called the centrifugal force. The word ‘centrifugal’ means ‘flee from center’.

Note that the ‘centrifugal force’ appears to act on the particle, only when we analyse the motion from a rotating frame. With respect to an inertial frame there is only centripetal force which is given by the tension in the rstring. For this reason centrifugal force is called as a ‘pseudo force’. A pseudo force has no origin. It arises due to the non inertial nature of the frame considered. When circular motion problems are solved from a rotating frame of reference, while drawing free body diagram of a particle, the centrifugal force should necessarily be included as shown in the figure.

Question 8.
Briefly explain ‘rolling friction’.
Answer:
The invention of the wheel plays a crucial role in human civilization. One of the important applications is suitcases with rolling on coasters. Rolling wheels makes it easier than carrying luggage. When an object moves on a surface, essentially it is sliding on it. But wheels move on the surface through rolling motion. In rolling motion when a wheel moves on a surface, the point of contact with surface is always at rest. Since Rolling and kinetic friction the point of contact is at rest, there is no relative motion between the wheel and surface.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Hence the frictional fore is very less. At the same time if an object moves without a wheel, there is a relative motion between the object and the surface. As a result frictional force is larger. This makes it difficult to move the object. The figure shows the difference between rolling and kinetic friction. Ideally in pure rolling, motion of the point of contact with the surface should be at rest, but in practice it is not so.

Due to the elastic nature of the surface at the point of contact there will be some deformation on the object at this point on the wheel or surface as shown in figure. Due to this deformation, there will be minimal friction between wheel and surface. It is called ‘rolling friction. In fact, rolling friction’ is much smaller than kinetic friction.

Question 9.
Describe the method of measuring angle of repose.
Answer:
When objects are connected by strings and When objects are connected by strings and a force F is applied either vertically or horizontally or along an inclined plane, it produces a tension T in the string, which affects the acceleration to an extent. Let us discuss various cases for the same.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Case 1:
Vertical motion:
Consider two blocks of masses m1 and m2 (m1> m2) connected by a light and in extensible string that passes over a pulley as shown in Figure.

Let the tension in the string be T and acceleration a. When the system is released, both the blocks start moving, m2 vertically upward and mk, downward with same acceleration a. The gravitational force m1g on mass m1 is used in lifting the mass m2. The upward direction is chosen as y direction. The free body diagrams of both masses are shown in Figure.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Applying Newton’s second law for mass m2
T \(\hat{j}\) – m2g\(\hat{j}\) = m2a\(\hat{j}\) The left hand side of the above equation is the total force that acts on m2 and the right hand side is the product of mass and acceleration of m2 in y direction.
By comparing the components on both sides, we get
T = m2g = m2a ……….(1)

Similarly, applying Newton’s second law for mass m2
T \(\hat{j}\) – m1g\(\hat{j}\) = m1a\(\hat{j}\)
As mass mj moves downward (-\(\hat{j}\)), its acceleration is along (-\(\hat{j}\))
By comparing the components on both sides, we get
T = m1g = -m1a
m1g – T = m1a ………..(2)

Adding equations (1) and (2), we get
m1g – m2g = m1a + m2a
(m1 – m2)g = (m1 + m2)a …………(3)
From equation (3), the acceleration of both the masses is –
a = (\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))g ………..(4)
If both the masses are equal (m1 = m2), from equation (4)
a = 0
This shows that if the masses are equal, there is no acceleration and the system as a whole will be at rest.
To find the tension acting on the string, substitute the acceleration from the equation (4) into the equation (1).
T = m2g = m2(\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))
T = m2g + m2 (\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))g ……….(5)
By taking m2g common in the RHS of equation (5)
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Equation (4) gives only magnitude of acceleration
For mass m1, the acceleration vector is given by \(\vec{a}\) = –\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\)\(\hat{j}\)
For mass m2, the acceleration vector is given by \(\vec{a}\) = \(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\) \(\hat{j}\)

Case 2:
Horizontal motion:
In this case, mass m2 is kept on a horizontal table and mass m1, is hanging through a small pulley as shown in figure. Assume that there is no friction on the surface
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
As both the blocks are connected to the un stretchable string, if m1 moves with an acceleration a downward then m2 also moves with the same acceleration a horizontally.
The forces acting on mass m2 are

  1. Downward gravitational force (m2g)
  2. Upward normal force (N) exerted by the surface
  3. Horizontal tension (T) exerted by the string

The forces acting on mass m1 are

  1. Downward gravitational force (m1g)
  2. Tension (T) acting upwards

The free body diagrams for both the masses is shown in figure 2.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Applying Newton’s second law for m1
T\(\hat{i}\) – m1g\(\hat{j}\) = -m1a\(\hat{j}\) (alongy direction)
By comparing the components on both sides of the above equation,
T – m1g = -m1a …………(1)
Applying Newton’s second law for m2
T\(\hat{i}\) = m1a\(\hat{i}\) (along x direction)
By comparing the components on both sides of above equation,
T = m2a ………….(2)
There is no acceleration along y direction for m2.
N\(\hat{j}\) – m2g\(\hat{j}\) = 0
By comparing the components on both sides of the above equation
N – m2g = 0
N = m2g ……….(3)
By substituting equation (2) in equation (1), we can find the tension T
m2a – m1g = -m1a
m2a + m1a = m1g
a = \(\frac{m_{1}}{m_{1}+m_{2}}\)g …………(4)
Tension in the string can be obtained by substituting equation (4) in equation (2)
T = \(\frac{m_{1} m_{2}}{m_{1}+m_{2}}\)g ………..(5)

Comparing motion in both cases, it is clear that the tension in the string for horizontal motion is half of the tension for vertical motion for same set of masses and strings. This result has an important application in industries. The ropes used in conveyor belts (horizontal motion) work for longer duration than those of cranes and lifts (vertical motion).

Question 10.
Explain the need for banking of tracks.
Answer:
In a leveled circular road, skidding mainly depends on the coefficient of static friction py. The coefficient of static friction depends on the nature of the surface which has a maximum limiting value. To avoid this problem, usually the outer edge of the road is slightly raised compared to inner edge. This is called banking of roads or tracks. This introduces an inclination, and the angle is called banking angle. “Let the surface of the road make angle 9 with horizontal surface. Then the normal force makes the same angle 9 with the vertical. When the car takes a turn.Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

there are two forces acting on the car:
(a) Gravitational force mg (downwards)
(b) Normal force N (perpendicular to surface)
We can resolve the normal force into two components N cos θ and N sin θ. The component balances the downward gravitational force ‘mg’ and component will provide the necessary centripetal acceleration. By using Newton second law.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

The banking angle 0 and radius of curvature of the road or track determines the safe speed of the car at the turning. If the speed of car exceeds this safe speed, then it starts to skid outward but frictional force comes into effect and provides an additional centripetal force to prevent the outward skidding.

At the same time, if the speed of the car is little lesser than safe speed, it starts to skid inward and frictional force comes into effect, which reduces centripetal force to prevent inward skidding. However if the speed of the vehicle is sufficiently greater than the correct speed, then frictional force cannot stop the car from skidding.

Question 11.
Calculate the centripetal acceleration of Moon towards the Earth.
Answer:
The centripetal acceleration is given by a = \(\frac{v^{2}}{r}\) This expression explicitly depends on Moon’s speed which is nontrivial. We can work with the formula
ω2Rm = am
am is centripetal acceleration of the Moon due to Earth’s gravity, ω is angular velocity
Rm is the distance between Earth and the Moon, which is 60 times the radius of the Earth.
Rm = 60R = 60 x 6.4 x 106 = 384 x 106 m
As we know the angular velocity ω = \(\frac { 2π}{ T }\) and T = 27.3 days = 27.3 x 24 x 60 x 60 second = 2.358 x 106 sec.
By substituting these values in the formula for acceleration
a6 = \(\frac{\left(4 \pi^{2}\right)\left(384 \times 10^{6}\right)}{\left(2.358 \times 10^{8}\right)^{2}}\) = 0.00272 ms-2

Samacheer Kalvi 11th Physics Laws of Motion Conceptual Questions

Question 1.
Why it is not possible to push a car from inside?
Answer:
While trying to push a car from outside, he pushes the ground backwards at an angle. The ground offers an equal reaction in the opposite direction, so car can be moved. But the person sits inside means car and the person becomes a single system, and the force given will be a internal force. According to Newton’s third law, total internal force acting on the system is zero and it cannot accelerate the system.

Question 2.
There is a limit beyond which the polishing of a surface increases frictional resistance rather than decreasing it why?
Answer:
Friction is a contact force. Friction is directly proportional to area of contact. In the normal surfaces there are bumps to interlock the surfaces in contact. But the surfaces are polished beyond certain limit. The area of contact will be increased and the molecules come closer to each other. It increases electrostatic force between the molecules. As a result it increases friction.

Question 3.
Can a single isolated force exist in nature? Explain your answer.
Answer:
No. According to Newton’s third law, for every action, there is an equal and opposite reaction. So, whatever case we consider, if there is an action there is always a reaction. So it is impossible.

Question 4.
Why does a parachute descend slowly?
Answer:
A parachute descends slowly because the surface area of parachute is large so that air gives more resistance when it descends down.

Question 5.
When walking on ice one should take short steps. Why?
Answer:
Let R represent the reaction offered by the ground. The vertical component R cos θ will balance the weight of the person and the horizontal component R sin θ will help the person to walk forward.
Now, normal reaction = R cos θ
Friction force = R sin θ
Coefficient of friction, µ = \(\frac {R sin θ}{R cos θ }\) = tan θ
In a long step, θ is more. So tan θ is more. But p has a fixed value. So, there is danger of slipping in a long step.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Question 6.
When a person walks on a surface, the frictional force exerted by the surface on the person is opposite to the direction of motion. True or false?
Answer:
False. In frictional force exerted by the surface on the person is in the direction of his motion. Frictional force acts as an external force to move the person. When the person trying to move, he gives a push to ground on the backward direction and by Newton’s third law he is pushed by the ground in the forward direction. Hence frictional force acts along the direction of motion.

Question 7.
Can the coefficient of friction be more than one?
Answer:
Yes. The coefficient of friction can be more than one in some cases such as silicone rubber. Coefficient of friction is the ratio of frictional force to normal force, i.e. F = μR. If p is greater than one means frictional force is greater than normal force. But in general case the value is usually between 0 and 1.

Question 8.
Can we predict the direction of motion of a body from the direction of force on it?
Answer:
Yes. The direction of motion is always opposite to the force of kinetic friction. By using the principle of equilibrium, the direction of force of static friction can be determined. When the object is in equilibrium, the frictional force must point in the direction which results as a net force is zero.

Question 9.
The momentum of a system of particles is always conserved. True or false?
Answer:
True. The total momentum of a system of particles is always constant i.e. conserved. When no external force acts on it.

Samacheer Kalvi 11th Physics Laws of Motion Numerical Problems

Question 1.
A force of 50 N act on the object of mass 20 kg. shown in the figure. Calculate the acceleration of the object in x and y directions.
Answer:
Given F = 50 N and m = 20 kg
(1) component of force along x – direction
Fx = F cos θ
= 50 x cos 30° = 43.30 N
ax = \(\frac{F_{x}}{m}\) = \(\frac {43.30}{20}\) =2.165 ms-2

(2) Component of force along y – direction
Fy = F sin θ = 50 sin 30° = 25 N
ay = \(\frac{F_{y}}{m}\) = \(\frac {25}{20}\) = 1.25 ms-2

Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Question 2.
A spider of mass 50 g is hanging on a string of a cob web as shown in the figure. What is the tension in the string?
Answer:
Given m = 50 g, T = ?
Tension in the string T = mg
= 50 x 10-2 x 9.8 = 0.49 N

Question 3.
What is the reading shown in spring balance?
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Answer:
When a spring balance hung on a rigid support and load is attached at its other end, the weight of the load exerts a force on the rigid support in turn support exerts equal and opposite force on that load, so that balance will be stretched. This is the principle of spring balance. Flence the answer is 4 kg.
Given: m = 2 kg, 0 = 30°.
Resolve the weight into its component as mg sin θ and mg cos θ.
Here mg sin θ acts parallel to the surface
∴ W = mg sin θ
= 2 x 9.8 x sin 30° = 2 x 9.8 x \(\frac {1}{2}\) = 9.8 N

Question 4.
The physics books are stacked on each other in the sequence: +1 volumes 1 and 2; +2 volumes 1 and 2 on a table
(a) Identify the forces acting on each book and draw the free body diagram.
(b) Identify the forces exerted by each book on the other.
Answer:
Let m1, m2, m3, m4, are the masses of +1 volume I and II and +2 volumes I & II
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

(a) Force on book m4

  • Downward gravitational force acting downward (m3g)
  • Upward normal force (N3) exerted by book of mass m3
    Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion |

(b) Force on book m3

  • Downward gravitational force (m3g)
  • Downward force exerted by m4 (N4)
  • Upward force exerted by m2 (N2)
    Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

(c) Force on book m2

  • Downward gravitational force (m2g)
  • Downward force exerted by m3 (N3)
  • Upward force exerted by m1 (CN1)
    Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

(d) Force on book m1

  • Downward gravitational force exerted by earth (m1g)
  • Downward force exerted by m2 (N2)
  • Upward force exerted by the table (Ntable)
    Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Question 5.
A bob attached to the string oscillates back and forth. Resolve the forces acting on the bob in to components. What is the acceleration experienced by the bob at an angle θ.
Answer:
The gravitational force (mg) acting downward can be resolved into two components as mg cos θ and mg sin θ
T – tension exerted by the string.
Tangential force FT = maT = mg sin θ
∴ Tangential acceleration aT = g sin θ
Centripetal force Fc = mac = T – mg cos θ
ac = \(\frac { T – mg cos θ }{ m }\)
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Question 6.
Two masses m1 and m2 are connected with a string passing over a friction-less pulley fixed at the comer of the table as shown in the figure. The coefficient of static friction of mass m1 with the table is µs Calculate the minimum mass m3 that may be placed on m1to prevent it from sliding. Check if m1 = 15 kg, m2 = 10 kg, m3 = 25 and µs = 0.2.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Solution:
Let m3 is the mass added on m1
Maximal static friction
\(f_{s}^{\max }\) = µsN = µs (m1 + m3 )g
Here
N = (m1 + m3 )g
Tension acting on string = T = m2 g
Equate (1) and (2)
µs(m1 + m3) = m2g
µsm1 + µsm3 = m2
m3 = \(f_{s}^{\max }\) – m1

(ii) Given,
m1 = 15 kg, m2 = 10 kg : m3 = 25 kg and µs = 0.2
m3 = \(f_{s}^{\max }\) – m1
m3 = \(\frac {10}{ 0.2 }\) – 15 = 50 – 15 = 35 kg
The minimum mass m3 = 35 kg has to be placed on ml to prevent it from sliding. But here m3 = 25 kg only.
The combined masses (m1 + m3) will slide.

Question 7.
Calculate the acceleration of the bicycle of mass 25 kg as shown in Figures 1 and 2.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Answer:
Given:
Mass of bicycle m = 25 kg
Fig. I:
Net force acting in the forward direction, F = 500 – 400 = 100 N
acceleration a = \(\frac { F}{ m }\) = \(\frac { 100 }{25}\) = 4 ms-2

Fig. II:
Net force acting on bicycle F = 400 – 400 = 0
∴ acceleration a = \(\frac { F}{ m }\) = \(\frac { 0}{25}\) = 0

Question 8.
Apply Lami’s theorem on sling shot and calculate the tension in each string?
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Answer:
Given F = 50 N, θ = 30°
Here T is resolved into its components as T sin θ and T cos θ as shown.
According to Lami’s theorem,
\(\frac { F}{sin θ}\) = \(\frac { T}{sin (180 – θ)}\) = \(\frac { T}{sin (180 – θ)}\)
\(\frac { F}{sin θ}\) = \(\frac { T}{sin θ}\)
\(\frac { F}{2 sin θ cos θ}\) = \(\frac { T}{sin θ}\) [T = \(\frac { T}{2 cos θ}\) ]
T = \(\frac {T}{2 cos θ}\) = \(\frac { 50}{ 2 cos 30}\) = 28.868 N

Question 9.
A football player kicks a 0.8 kg ball and imparts it a velocity 12 ms-1. The contact between the foot and ball is only for one – sixtieth of a second. Find the average kicking force.
Answer:
Given,
Mass of the ball = 0.8 kg
Final velocity (V) =12 ms-1 and time t =\(\frac {1}{60}\) s
Initial velocity = 0
We know the average kicking force
F = ma = \(\frac {m(v – u)}{t}\) = \(\frac{0.8(12-0)}{\left(\frac{1}{60}\right)}\)
F = 576 N

Question 10.
A stone of mass 2 kg is attached to a string of length 1 meter. The string can withstand maximum tension 200 N. What is the maximum speed that stone can have during the whirling motion?
Solution:
Given,
Mass of a stone = 2 kg,
length of a string = 1 m
Maximum tension = 200 N
The force acting on a stone in the whirling motion is centripetal force. Which is provided by tension of the string.
Tmax = Fmax = \(\frac{m \mathrm{V}_{\mathrm{max}}^{2}}{r}\)
200 = \(v_{\max }^{2}\) = 100
\(v_{\max }^{2}\) = 10 ms-1

Question 11.
Imagine that the gravitational force between Earth and Moon is provided by an invisible string that exists between the Moon and Earth. What is the tension that exists in this invisible string due’ to Earth’s centripetal force? (Mass of the Moon = 7.34 x 1022 kg, Distance between Moon and Earth = 3.84 x 108 m).

Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Solution:
Given,
Mass of the moon = 7.34 x 1022 kg
Distance between moon and earth = 3.84 x 108 m
Centripetal force = F = \(\frac{m \mathrm{V}^{2}}{r}\) = \(\frac{7.34 \times 10^{22} \times\left(1.023 \times 10^{3}\right)^{2}}{3.84 \times 10^{8}}\) = 2 x 1020

Question 12.
Two bodies of masses 15 kg and 10 kg are connected with light string kept on a smooth surface. A horizontal force F = 500 N is applied to a 15 kg as shown in the figure. Calculate the tension acting in the string.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Answer:
Given,
m1  = 15 kg, m2  = 10 kg, F = 500 N
Tension acting in the string T = \(\frac{m_{2}}{m_{1}+m_{2}}\) F
T = \(\frac {10}{25}\) x 500 = 200 N

Question 13.
People often say “For every action there is an equivalent opposite reaction”. Here they meant ‘action of a human’. Is it correct to apply Newton’s third law to human actions? What is meant by ‘action’ in Newton third law? Give your arguments based on Newton’s laws.
Answer:
Newton’s third law is applicable to only human’s physical actions which involves physical force. Third law is not applicable to human’s psychological actions or thoughts.

Question 14.
A car takes a turn with velocity 50 ms-1 on the circular road of radius of curvature To m. Calculate the centrifugal force experienced by a person of mass 60 kg inside the car?
Answer:
Given,
Mass of a person = 60 kg
Velocity of the car = 50 ms-1
Radius of curvature = 10 m
Centrifugal force F = \(\frac{m \mathrm{V}^{2}}{r}\) = \(\frac{60 \times(50)^{2}}{10}\) = 15,000 N

Question 15.
A long stick rests on the surface. A person standing 10 m away from the stick. With what minimum speed an object of mass 0.5 kg should he thrown so that it hits the stick. (Assume the coefficient of kinetic friction is 0.7).
Answer:
Given,
Distance (s) = 10 m
Mass of the object (m) = 0.5 kg
Coefficient of kinetic friction (µ) = 0.7
Work done in moving a body in horizontal surface ω = µR x s = µmg x s
This work done is equal to initial kinetic energy of the object
\(\frac{1}{2} m v^{2}\) = µ mg s
\(\left|v^{2}\right|\) = 2 µgs = 2 x 0.7 x 9.8 x 10
v2 = 14 x 9.8 = 137. 2
v = 11. 71 ms-1

Samacheer Kalvi 11th Physics Laws of Motion Additional Questions Solved

Samacheer Kalvi 11th Physics Laws of Motion Multiple Choice Questions

Question 1.
The concept “force causes motion” was given by –
(a) Galileo
(b) Aristotle
(c) Newton
(d) Joule
Answer:
(b) Aristotle

Question 2.
Who decoupled the motion and force?
(a) Galileo
(b) Aristotle
(c) Newton
(d) Joule
Answer:
(a) Galileo

Question 3.
The inability of objects to move on its own or change its state of motion is called as –
(a) force
(b) momentum
(c) inertia
(d) impulse
Answer:
(c) inertia

Question 4.
Inertia means –
(a) inability
(b) resistance to change its state
(c) movement
(d) inertial frame
Answer:
(b) resistance to change its state

Question 5.
When a bus starts to move from rest, the passengers experience a sudden backward push is an example for –
(a) Inertia of motion
(b) Inertia of direction
(c) Inertia of rest
(d) back pull
Answer:
(c) Inertia of rest

Question 6.
If the brake is applied in the moving bus suddenly, passengers move forward is an example for –
(a) Inertia of motion
(b) Inertia of direction
(c) Inertia of rest
(d) back pull
Answer:
(a) Inertia of motion

Question 7.
In whirling motion, if the string is cut suddenly, the stone moves tangential to circle is an –
(a) Inertia of motion
(b) Inertia of direction
(c) Inertia of rest
(d) back pull
Answer:
(b) Inertia of direction

Question 8.
Newtons laws are applicable in –
(a) Inertial frame
(b) non inertial frame
(c) in any frame
(d) none
Answer:
(a) Inertial frame

Question 9.
The accelerated train is an example for –
(a) inertial frame
(b) non-inertial frame
(c) both (a) and (b)
(d) none of the above
Answer:
(b) non-inertial frame

Question 10.
Rate of change of momentum of an object is equal to –
(a) acceleration
(b) work done
(c) force
(d) impulse
Answer:
(c) force

Question 11.
The product of mass and velocity is –
(a) force
(b) impulse
(c) momentum
(d) acceleration
Answer:
(c) momentum

Question 12.
Unit of momentum –
(a) kg ms-2
(b) kg ms-1
(c) MLT-2
(d) MLT-1
Answer:
(b) kg ms-1

Question 13.
According to Newton’s third law –
(a) F12 = F21
(*) F12 = -F21
(c) F12 + F21 = 0
(d) F12 x F21 = 0
Answer:
(a) F12 = F21

Question 14.
According to Newton’s third law –
(a) \(\overrightarrow{\mathrm{F}_{12}}=\overrightarrow{\mathrm{F}_{21}}\)
(b) \(\overrightarrow{\mathrm{F}_{12}}=-\overrightarrow{\mathrm{F}_{21}}\)
(c) \(\mathrm{F}_{12}+\mathrm{F}_{21}\) = 0
(d) \(\mathrm{F}_{12}x\mathrm{F}_{21}\) = 0
Answer:
(b) \(\overrightarrow{\mathrm{F}_{12}}=-\overrightarrow{\mathrm{F}_{21}}\)

Question 15.
The law which is valid in both inertial and non-inertial frame is –
(a) Newton’s first law
(b) Newton’s second law
(c) Newton’s third law
(d) none
Answer:
(c) Newton’s third law

Question 16.
When a force is applied on a body, it can change –
(a) velocity
(b) momentum
(c) direction of motion
(d) all the above
Answer:
(d) all the above

Question 17.
The rate of change of velocity is 1 ms-2 when a force is applied on the body of mass 75 gm the force is –
(a) 75 N
(b) 0.75 N
(c) 0.075 N
(d) 0.75 x 10-3 N
Answer:
(c) Force is given by
F = m a
= 75 gm x 1 cm s-2 = 75 x 10-3 x 1 = 75 x 10-3 = 0.075 N

Question 18.
The action and reaction forces acting on –
(a) same body
(b) different bodies
(c) either same or different bodies
(d) none of the above
Answer:
(b) different bodies

Question 19.
Newton’s first law of motion gives the concept of –
(a) velocity
(b) energy
(c) momentum
(d) Inertia
Answer:
(d) Inertia

Question 20.
Inertia of a body has direct dependence on –
(a) velocity
(b) area
(c) mass
(d) volume
Answer:
(c) mass

Question 21.
If a car and a scooter have the same momentum, then which one is having greater speed?
(a) scooter
(b) car
(c) both have same velocity
(d) data insufficient
Answer:
(a) scooter

Question 22.
Newton’s second law gives –
(a) \(\overrightarrow{\mathrm{F}} \propto \frac{d \overrightarrow{\mathrm{P}}}{\mathrm{dt}}\)
(b) \(\overrightarrow{\mathrm{F}}=\frac{d \overrightarrow{\mathrm{P}}}{\mathrm{dt}}\)
(c) \(\overrightarrow{\mathrm{F}}=m \vec{a}\)
(d) all the above
Answer:
(d) all the above

Question 23.
1 dyne is –
(a) 105N
(b) 10-5N
(c) 1N
(d) 10-3N
Answer:
(b) 10-5N

Question 24.
If same force is acting on two masses m1 and m2, and the accelerations of two bodies are a1 and a2 respectively, then –
(a) \(\frac{a_{2}}{a_{1}}=\frac{m_{2}}{m_{1}}\)
(b) \(\frac{a_{1}}{a_{2}}=\frac{m_{1}}{m_{2}}\)
(c) \(\frac{a_{1}}{a_{2}}=\frac{m_{2}}{m_{1}}\)
(d) m1 a1 + m2a2 = 0
Answer:
(c) \(\frac{a_{1}}{a_{2}}=\frac{m_{2}}{m_{1}}\)

Question 25.
If a force \(\overline{\mathrm{F}}\) = 3\(\hat{i}\) – 4\(\hat{j}\) N produces an acceleration of 10 ms-2 on a body, then the mass of a body is –
(a) 10 kg
(b) 9 kg
(c) 0.9 kg
(d) 0.5 kg
Answer:
\(\overline{\mathrm{F}}\) = 3\(\hat{i}\) – 4\(\hat{j}\)
Magnitude:
|\(\overline{\mathrm{F}}\)| = \(\sqrt{9+16}\) = \(\sqrt{25}\) = 5N
F = ma
⇒ m = \(\frac{|\mathrm{F}|}{a}\) = \(\frac{5}{10}\) = \(\frac{1}{2}\) = 0.5 kg

Question 26.
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms-1. How long does the body take to stop?
(a) 0.75 s
(b) 1.33 s
(c) 6 s
(d) 35 s
Answer:
Acceleration a = \(\frac{-F}{m}\) = \(\frac{50}{20}\) = – 2.5 ms-2
u = l5 ms-1
v = 0
t = ?
v = u + at
0 = 15 – 2.5t
t = \(\frac{15}{2.5}\) = 6s

Question 27.
Rain drops come down with –
(a) zero acceleration and non zero velocity
(b) zero velocity with non zero acceleration
(c) zero acceleration and non zero net force
(d) none
Answer:
(a) zero acceleration and non zero velocity

Question 28.
If force is the cause then the effect is –
(a) mass
(b) potential energy
(c) acceleration
(d) Inertia
Answer:
(c) acceleration

Question 29.
In free body diagram, the object is represented by a –
(a) line
(b) arrow
(c) circle
(d) point
Answer:
(d) point

Question 30.
When an object of mass m slides on a friction less surface inclined at an angle 0, then normal force exerted by the surface is –
(a) g cos θ
(b) mg cos θ
(c) g sin θ
(d) mg tan θ
Answer:
(b) mg cos θ

Question 31.
The acceleration of the sliding object in an inclined plane –
(a) g cos θ
(b) mg cos θ
(c) g sin θ
(d) mg sin θ
Answer:
(c) g sin θ

Question 32.
The speed of an object sliding in an inclined plane at the bottom is –
(a) mg cos θ
(b) \(\sqrt{2 s g sin θ}\)
(c) \(\sqrt{2 s g cos θ}\)
(d) \(\sqrt{2 s g tan θ}\)
Answer:
(b) \(\sqrt{2 s g sin θ}\)

Question 33.
The acceleration of two bodies of mass m1 and m2 in contact on a horizontal surface is –
(a) \(a=\frac{\mathbf{F}}{m_{1}}\)
(b) \(a=\frac{F}{m_{2}}\)
(c)  \(a=\frac{\mathrm{F}}{m_{1}+m_{2}}\)
(d) \(a=\frac{\mathrm{F}}{m_{1} m_{2}}\)
Answer:
(c)  \(a=\frac{\mathrm{F}}{m_{1}+m_{2}}\)

Question 34.
Two blocks of masses m1 and m2 (m1 > m2) in contact with each other on frictionless, horizontal surface. If a horizontal force F is given on m1, set into motion with acceleration a, then reaction force on mass m1 by m2, is –
(a) \(\frac{\mathrm{F} m_{1}}{m_{1}+m_{2}}\)
(b) \(\frac{m_{1} m_{2}}{\mathrm{F} m_{1}}\)
(c) \(\frac{m_{1} m_{2}}{\mathrm{F} m_{2}}\)
(d) \(\frac{\mathrm{F} m_{2}}{m_{1}+m_{2}}\)
Answer:
(d) \(\frac{\mathrm{F} m_{2}}{m_{1}+m_{2}}\)

Question 35.
If two masses m1 and m2 (m1 > m2) tied to string moving over a frictionless pulley, then acceleration of masses –
(a) \(\frac{\left(m_{1}-m_{2}\right)}{m_{1}+m_{2}}\) g
(b) \(\frac{m_{1}+m_{2}}{\left(m_{1}-m_{2}\right)}\) g
(c) \(\frac{2 m_{1} m_{2}}{m_{1}+m_{2}}\) g
(d) \(\frac{m_{1} m_{2}}{2 m_{1} m_{2}}\) g
Answer:
(a) \(\frac{\left(m_{1}-m_{2}\right)}{m_{1}+m_{2}}\) g

Question 36.
if two masses m1 and m2 (m1 > m2)tied to string moving over a frictionless pulley, then acceleration of masses –
(a) \(\frac{\left(m_{1}-m_{2}\right)}{m_{1}+m_{2}}\) g
(b) \(\frac{m_{1}+m_{2}}{\left(m_{1}-m_{2}\right)}\) g
(c) \(\frac{2 m_{1} m_{2}}{m_{1}+m_{2}}\) g
(d) \(\frac{m_{1} m_{2}}{2 m_{1} m_{2}}\) g
Answer:
(a) \(\frac{\left(m_{1}-m_{2}\right)}{m_{1}+m_{2}}\) g

Question 37.
Three massses is in contact as shown. If force F is applied to mass m1, the acceleration of three masses is –
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
(b) \(\frac{m_{1} F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(c) \(\frac{\left(m_{2}+m_{3}\right) F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(d) \(\frac{m_{3} \mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion q37
Answer:
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}+m_{3}}\)

Question 38.
Three masses in contact is as shown above. If force F is applied to mass m1 then the contact force acting on mass m2 is –
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
(b) \(\frac{m_{1} F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(c) \(\frac{\left(m_{2}+m_{3}\right) F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(d) \(\frac{m_{3} \mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
Answer:
(c) \(\frac{\left(m_{2}+m_{3}\right) F}{\left(m_{1}+m_{2}+m_{3}\right)}\)

Question 39.
Three masses is contact as shown. It force F is applied to mass m1, then the contact force acting on mass m3 is –
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
(b) \(\frac{m_{1} F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(c) \(\frac{\left(m_{2}+m_{3}\right) F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(d) \(\frac{m_{3} \mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
Answer:
(d) \(\frac{m_{3} \mathrm{F}}{m_{1}+m_{2}+m_{3}}\)

Question 40.
Two masses connected with a string. When a force F is applied on mass m2. The acceleration produced is –
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion q40
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}}\)
(b) \(\frac{\mathbf{F}}{m_{1}-m_{2}}\)
(c) \(\frac{m_{1}+m_{2}}{\mathrm{F}}\)
(d) \(\frac{m_{3} \mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
Answer:
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}}\)

Question 41.
Two masses connected with a string. When a force F is applied on mass m2. The force acting on m1 is –
(a) \(\frac{m_{1} \mathrm{F}}{m_{1}+m_{2}}\)
(b) \(\frac{m_{2} \mathrm{F}}{m_{1}+m_{2}}\)
(c) \(\frac{m_{1}+m_{2}}{m_{1}} \mathbf{F}\)
(d) \(\frac{m_{1}+m_{2}}{m_{2}} \mathbf{F}\)
Answer:
(b) \(\frac{m_{2} \mathrm{F}}{m_{1}+m_{2}}\)

Question 42.
If a block of mass m lying on a frictionless inclined plane of length L height h and angle of inclination θ, then the velocity at its bottom is –
(a) g sin θ
(b) g cos θ
(c) \(\sqrt{2 g h}\)
(d) \(\sqrt{2 a sin θ}\)
Answer:
(c) \(\sqrt{2 g h}\)

Question 43.
If a block of mass m lying on a frictionless inclined plane of length L, height h and angle of inclination θ, then the time take taken to reach the bottom is –
(a) g sing θ
(b) sin θ \(\sqrt{\frac{2 h}{g}}\)
(c) sin θ \(\sqrt{\frac{g}{h}}\)
(d) \(\frac{1}{\sin \theta} \sqrt{\frac{2 h}{g}}\)
Answer:
(d) \(\frac{1}{\sin \theta} \sqrt{\frac{2 h}{g}}\)

Question 44.
A rocket works on the principle of conservation of –
(a) energy
(b) mass
(c) angular momentum
(d) linear momentum
Answer:
(b) mass

Question 45.
A bomb at rest explodes. The total momentum of all its fragments is –
(a) zero
(b) infinity
(c) always 1
(d) always greater then 1
Answer:
(a) zero

Question 46.
A block of mass m1 is pulled along a horizontal friction-less surface by a rope of mass m2 If a force F is given at its free end. The net force acting on the block is –
(a) \(\frac{m_{1} \mathrm{F}}{m_{1}-m_{2}}\)
(b) F
(c) \(\frac{m_{2} \mathrm{F}}{\left(m_{1}+m_{2}\right)}\)
(d) \(\frac{m_{1} \mathrm{F}}{\left(m_{1}+m_{2}\right)}\)
Answer:
(b) F

Question 47.
A block of mass m is pulled along a horizontal surface by a rope. The tension in the rope will be same at all the points –
(a) if the rope is accelerated
(b) if the rope is mass less
(c) always
(d) none of the above
Answer:
(b) if the rope is mass less

Question 48.
The lines of forces act at a common point is called as –
(a) concurrent forces
(b) co-planar forces
(c) equilibrium
(d) resultant
Answer:
(a) concurrent forces

Question 49.
If the lines of forces act in the same plane, they can be –
(a) concurrent forces
(b) coplanar forces
(c) either concurrent force or coplanar forces
(d) Lami’s force
Answer:
(d) concurrent forces

Question 50.
Lami’s theorem is applicable only when the system of forces are is –
(a) same plane
(b) different plane
(c) equilibrium
(d) none of the above
Answer:
(c) equilibrium

Question 51.
Due to the action of internal forces of the system, the total linear momentum of the system is –
(a) a variable
(b) a constant
(c) always zero
(d) always infinity
Answer:
(c) always zero

Question 52.
The velocity with which a gun suddenly moves backward after firing is –
(a) linear velocity
(b) positive velocity
(c) recoil velocity
(d) v1 + v2
Answer:
(c) recoil velocity

Question 53.
If a very large force acts on an object for a very short duration, then the force is called as –
(a) Newtonian force
(b) impulsive force
(c) concurrent force
(d) coplanar force
Answer:
(A) impulsive force

Question 54.
The unit of impulse is –
(a) Nm
(b) Ns
(c) Nm2
(d) Ns-2
Answer:
(b) Ns

Question 55.
The force which always opposes the relative motion between an object and the surface where it is placed is –
(a) concurrent force
(b) frictional force
(c) impulsive force
(d) coplanar force
Answer:
(b) frictional force

Question 56.
The force which opposes the initiation of motion of an object on the surface is –
(a) static friction
(b) kinetic friction
(c) friction
(d) zero
Answer:
(d) static friction

Question 57.
When the object is at rest, the resultant of gravitational force and upward normal force is –
(a) Static force
(b) zero
(c) one
(d) infinity
Answer:
(b) zero

Question 58.
The magnitude of static frictional force d lies between –
(a) 0 ≤ f ≤ µsN
(b) 0 ≥f ≥ µsN
(c) 0 and 1
(d) 0 and minimal static frictional force.
Answer:
(a) 0 ≤ f ≤ µsN

Question 59.
The unit of co-efficient of static friction is –
(a) N
(b) N m
(c) N s
(d) no unit
Answer:
(d) no unit

Question 60.
If the object is at rest and no external force is applied on the object, the static friction acting on the object is –
(a) µsN
(b) zero
(c) one
(d) infinity
Answer:
(d) no unit

Question 61.
When object begins to slide, the static friction acting on the object attains –
(a) zero
(b) minimum
(c) maximum
(d) infinity
Answer:
(c) maximum

Question 62.
The static friction does not depend upon –
(a) the area of contact
(b) normal force
(c) the magnitude of applied force
(d) none of the above
Answer:
(a) the area of contact

Question 63.
Which of the following pairs of materials has minimum amount of coefficient of static friction is –
(a) Glass and glass
(b) wood and wood
(c) ice and ice
(d) steel and steel
Answer:
(c) ice and ice

Question 64.
Kinetic friction is also called as –
(a) sliding friction
(b) dynamic friction
(c) both (a) and (b)
(d) static friction
Answer:
(c) both (a) and (b)

Question 65.
The unit of coefficient of kinetic friction is/has –
(a) Nm
(b) Ns
(c) Nm2
(d) no unit
Answer:
(d) no unit

Question 66.
The nature of materials in mutual contact decides –
(a) µs
(b) µk
(c) µs or µk
(d) none
Answer:
(c) µs or µk

Question 67.
Coefficient of kinetic friction is less than –
(a) O
(b) one
(c) µs
(d) µsN
Answer:
(c) µs

Question 68.
The static friction –
(a) increases linearly
(b) is constant
(c) zero
(d) varies parabolically
Answer:
(a) increases linearly

Question 69.
The kinetic friction –
(a) increases linearly
(b) is constant
(c) zero
(d) varies parabolically
Answer:
(b) is constant

Question 70.
Kinetic friction is independent of –
(a) nature of materials
(b) temperature of the surface
(c) applied force
(d) none of the above
Answer:
(c) applied force

Question 71.
The angle between the normal force and the resultant force of normal force and maximum frictional force is –
(a) angle of friction
(b) angle of repose
(c) angle of inclination
(d) none of the above
Answer:
(a) angle of friction

Question 72.
The angle friction θ is given by –
(a) tan µs
(b) tan-1 µs
(c) \(\frac{f S^{\mathrm{max}}}{N}\)
(d) sin-1 µs
Answer:
(b) tan-1 µs

Question 73.
The angle of inclined plane with the horizontal such that an object placed on it begins to slide is –
(a) angle of friction
(b) angle of repose
(c) angle of response
(d) angle of retardation
Answer:
(b) angle of repose

Question 74.
Comparatively, which of the following has lesser value than others?
(a) static friction
(b) kinetic friction
(c) Rolling friction
(d) skiping friction
Answer:
(c) Rolling friction

Question 75.
The origin of friction is –
(a) electrostatic interaction
(b) electromagnetic interaction magnetic
(c) photon interaction
(d) interaction
Answer:
(b) electromagnetic interaction

Question 76.
Friction can be reduced by –
(a) polishing
(b) lubricating
(c) using ball bearings
(d) all the above
Answer:
(c) using ball bearings

Question 77.
For a particle revolving in a circular path, the acceleration of the particle is –
(a) along the tangent
(b) along the radius
(c) along the circumference of the circle
(d) zero
Answer:
(b) along the radius

Question 78.
A particle moves along a circular path under the action of a force. The work done by the force is –
(a) Positive and non zero
(b) zero
(c) Negative and non zero
(d) none of the above
Answer:
(b) zero

Question 79.
A bullet hits and gets embedded in a solid block resting on a horizontal frictionless table. Which of the following is conserved?
(a) Momentum and kinetic energy
(b) kinetic energy alone
(c) Momentum alone
(d) potential energy alone
Answer:
(c) Momentum alone

Question 80.
The origin of the centripetal force can be –
(a) gravitational force
(b) frictional force
(c) coulomb force
(d) all the above
Answer:
(d) all the above

Question 81.
Centripetal acceleration is –
(a) \(\frac{m v^{2}}{r}\)
(b) \(\frac{v^{2}}{r}\)
(c) r v2
(d) rω
Answer:
(b) \(\frac{v^{2}}{r}\)

Question 82.
Centripetal acceleration is –
(a) \(\frac{m v^{2}}{r}\)
(b) r ω2
(c) rv2
(d) rω
Answer:
(c) rω2

Question 83.
The centripetal force is –
(a) \(\frac{m v^{2}}{r}\)
(b) rω2
(c) both (a) and (b)
(d) none
Answer:
(c) both (a) and (b)

Question 84.
When a car is moving on a circular track the centripetal force is due to –
(a) gravitational force
(b) frictional force
(c) magnetic force
(d) elastic force
Answer:
(b) frictional force

Question 85.
If the road is horizontal then the normal force and gravitational force are –
(a) equal and along the same direction
(b) equal and opposite
(c) unequal and along the same direction
(d) unequal and opposite
Answer:
(b) equal and opposite

Question 86.
The velocity of a car for safe turn on leveled circular road –
(a) \(v \leq \sqrt{\mu_{s} r g}\)
(b) \(v \geq \sqrt{\mu_{s} r g}\)
(c) \(v=\sqrt{\mu_{s} rg}\)
(d) \(v \leq \mu_{s} rg\)
Answer:
(a) \(v \leq \sqrt{\mu_{s} r g}\)

Question 87.
In a leveled circular road, skidding mainly depends on –
(a) µs
(b) µk
(c) acceleration
(d) none
Answer:
(a) µs

Question 88.
The speed of a car to move on the banked road so that it will have safe turn is –
(a) µsrg
(b) \(\sqrt{r g \tan \theta}\)
(c) rg tan θ
(d) r2g tan θ
Answer:
(b) \(\sqrt{r g \tan \theta}\)

Question 89.
Centrifugal force is a –
(a) pseudo force
(b) real force
(c) forced acting towards center
(d) none of the above
Answer:
(a) pseudo force

Question 90.
Origin of centrifugal force is due to –
(a) interaction between two
(b) inertia
(c) electromagnetic interaction
(d) inertial frame
Answer:
(b) inertia

Question 91.
Centripetal force acts in –
(a) inertial frame
(b) non inertial frame
(c) both (a) and (h)
(d) linear motion
Answer:
(c) both (a) and (b)

Question 92.
Centrifugal force acts in –
(a) inertial frame
(b) non inertial frame
(c) both (a) and (b)
(d) linear motion
Answer:
(b) non inertial frame

Question 93.
A cricket ball of mass loo g moving with a velocity of 20 ms-1 is brought to rest by a player in 0.05s the impulse of the ball is –
(a) 5 Ns
(b) – 2 Ns
(c) – 2.5 Ns
(d) zero
Answer:
(b) – 2 Ns
mass = 0.1 kg
Initial velocity t = 20 ms-1
Final velocity y = 0
Change in momentum in impulse = m(v – u) = 0.1(0 – 20) = – 2 Ns

Question 94.
If a stone tied at the one end of a string of length 0.5 m is whirled in a horizontal circle with a constant speed 6 ms-1  then the acceleration of the shone is –
(a) 12 ms-2
(b) 36 ms-2
(c) 2π2 ms-2
(d) 72 ms-2
Answer:
(d) Centripetal acceleration = \(\frac{v^{2}}{r}\) = \(\frac{6^{2}}{0.5}\) = \(\frac{36}{0.5}\) = 72 ms-2

Question 95.
A block of mass 3 kg is at rest on a rough inclined plane with angle of inclination 30° with horizontal. If .is 0.7, then the frictional force is –
(a) 17.82 N
(b) 1.81 N
(c) 3.63 N
(d) 2.1 N
Answer:
(a) Frictional force = µmg cos θ = 0.7 x 3 x 9.8 cos 30° = 17.82 N

Question 96.
Two masses 2 kg and 4 kg are tied at the ends of a mass less string and which is passing over a friction-less pulley. The tension in the string is –
(a) 3.68 N
(b) 78.4 N
(c) 26 N
(d) 13.26 N
Answer:
(c) Tension in the string T = \(\frac{2 m_{1} m_{2}}{\left(m_{1}+m_{2}\right)}\)g
T = \(\frac{2 x 2 x 4}{2 + 4}\) x 9.8 = \(\frac{16}{6}\) x 9.8 = 26.13 N

Question 97.
A bomb of 10 kg at rest explodes into two pieces of mass 4 kg and 6 kg. if the velocity of 4 kg mass is 6 ms-1 then the velocity of 6 kg is –
(a) – 4 ms-1
(b) – 6 ms-1
(c) – 24 ms-1
(d) – 2.2 ms-1
Answer:
(a) According to law of conservation of momentum
m1v1 + m2v2 = 0
v2 = –\(\frac{m_{1} v_{1}}{m_{2}}\) = \(\frac{4 x 6}{6}\) = -4ms-1

Question 98.
A body is subjected under three concurrent forces and it is in equilibrium. The resultant of any two forces is –
(a) coplanar with the third force
(b) is equal and opposite to third force
(c) both (a) and (b)
(d) none of the above
Answer:
(c) both (a) and (b)

Question 99.
An impulse is applied to a moving object with the force at an angle of 20° with respect to velocity vector. The angle between the impulse vector and the change in momentum vector is –
(a) 0°
(b) 30°
(c) 60°
(d) 120°
Answer:
(a) Impulse and change in momentum are in same direction. So the angle is zero.

Question 100.
A bullet of mass m and velocity v1 is fired into a large block of wood of mass M. The final velocity of the system is-
(a) \(\frac{v_{1}}{m+\mathrm{M}}\)
(b) \(\frac{m v_{1}}{m+\mathrm{M}}\)
(c) \(\frac{m+m}{m} v_{1}\)
(d) \(\frac{m+m}{m-M} v_{1}\)
Answer:
(b) \(\frac{m v_{1}}{m+\mathrm{M}}\)

Question 101.
A block of mass 2 kg is placed on the floor. The co – efficient of static friction is 0.4. The force of friction between the block and floor is –
(a) 2.8 N
(b) 7.8 N
(c) 2 N
(d) zero
Answer:
(b) The force required to move = = µR = µmg = 0.4 x 2 x 9.8 = 7.84 N

Question 102.
A truck weighing 1000 kg is moving with velocity of 50 km/h on smooth horizontal roads. A mass of 250 kg is dropped into it. The velocity with which it moves now is –
(a) 12.5 km/h
(b) 20 km/h
(c) 40 km/h
(d) 50 km/h
Answer:
(c) According to law of conservation of linear momentum
m2 v2 = (m1 + m2)v2
v2 = \(\frac{m_{1} v_{1}}{m_{1}+m_{2}}\) = \(\frac{1000 \times 50}{1250}\) = 40 km/h

Question 103.
A body of mass loo g is sliding from an inclined plane of inclination 30°. if u = 1.7, then the frictional force experienced is –
(a) \(\frac{3.4}{\sqrt{3}}\)N
(b) 1.47 N
(c) \(\frac{\sqrt{3}}{3.4}\)N
(d) 1.38 N
Answer:
(b) Frictional force F = µ mg cos θ = 1.7 x 0.1 x 10 cos 30°= \(\frac{1.7}{2}\) x \(\sqrt{3}\) = 1.47 N

Samacheer Kalvi 11th Physics Laws of Motion Short Answer Questions (1 Mark)

Question 1.
A passenger sitting in a car at rest, pushes the car from within. The car doesn’t move, why?
Answer:
For motion, there should be external force.

Question 2.
Give the magnitude and directions of the net force acting on a rain drop falling with a constant speed.
Answer:
as \(\overline{\mathrm{a}}\) = 0 so \(\overline{\mathrm{F}}\) = 0.

Question 3.
Why the passengers in a moving car are thrown outwards when it suddenly takes a turn?
Answer:
Due to inertia of direction.

Question 4.
You accelerate your car forward. What is the direction of the frictional force on a package resting on the floor of the car?
Answer:
The package in the accelerated car (a non inertial frame) experiences a Pseudo force in a direction opposite to that of the motion of the car. The frictional force on the package which acts opposite to this pseudo force is thus in the same direction (forward) as that of the car.

Question 5.
What is the purpose of using shockers in a car?
Answer:
To decrease the impact of force by increasing the time for which force acts.

Question 6.
Why are types made of rubber not of steel?
Answer:
Since coefficient of friction between rubber and road is less than the coefficient of friction between steel and road.

Question 7.
Wheels are made circular. Why?
Answer:
Rolling friction is less than sliding friction.

Question 8.
If a ball is thrown up in a moving train, it comes back to the thrower’s hands. Why?
Answer:
Both during its upward and downward motion, the ball continues to move inertia of motion with the same horizontal velocity as the train. In this period, the ball covers the same horizontal distance as the train and so it comes back to the thrower’s hand.

Question 9.
Calculate the force acting on a body which changes the momentum of the body at the rate of 1 kg-m/s2 .
Answer:
As F = rate change of momentum
F = 1 kg-m/s2 = 1N

Question 10.
On a rainy day skidding takes place along a curved path. Why?
Answer:
As the friction between the types and road reduces on a rainy day.

Question 11.
Why does a gun recoils when a bullet is being fired?
Answer:
To conserve momentum.

Question 12.
Why is it difficult to catch a cricket ball than a tennis ball even when both are moving with the same velocity?
Answer:
Being heavier, cricket ball has higher rate of change of momentum during motion so more force sumed.

Question 13.
The distance travelled by a moving body is directly proportional to time. Is any external force acting on it?
Answer:
As s ∝ t, so acceleration a = 0, therefore, no external force is acting on the body.

Question 14.
Calculate the impulse necessary to stop a 1500 kg car moving at a speed of 25 ms-1.
Answer:
Use formula I = change in momentum = m(v – u) (Impulse – 37500 Ns)

Question 15.
Lubricants are used between the two parts of a machine. Why?
Answer:
To reduce friction and so to reduce wear and tear.

Question 16.
What provides the centripetal force to a car taking a turn on a level road?
Answer:
Force of friction between the type and road provides centripetal force.

Question 17.
A body is acted upon by a number of external forces. Can it remain at rest?
Answer:
Yes, if the external forces acting on the body can be represented in magnitude and direction by the sides of a closed polygon taken in the same order.

Question 18.
Bodies of larger mass need greater initial effort to put them in motion. Why?
Answer:
As F = ma so for given a, more force will be required to put a large mass in motion.

Question 19.
An athlete runs a certain distance before taking a long jump Why?
Answer:
So that inertia of motion may help him in his muscular efforts to take a longer jump.

Question 20.
Action and reaction forces do not balance each other. Why?
Answer:
As they acts on different bodies.

Question 21.
The wheels of vehicles are provided with mudguards. Why?
Answer:
When the wheel rotates at a high speed, the mud sticking to the wheel flies off tangentially, this is due to inertia of direction. If order that the flying mud does not spoil the clothes of passer by the wheels are provided with mudguards.

Question 22.
China wares are wrapped in straw paper before packing. Why?
Answer:
The straw paper between the China ware increases the Time of experiencing the jerk during transportation. Hence impact of force reduces on China wares.

Question 23.
Why is it difficult to walk on a sand?
Answer:
Less reaction force.

Question 24.
The outer edge of a curved road is generally raised over the inner edge Why?
Answer:
In addition to the frictional force, a component of reaction force also provides centripetal force.

Question 25.
Explain why the water doesn’t fall even at the top of the circle when the bucket full of water is upside down rotating in a vertical circle?
Answer:
Weight of the water and bucket is used up in providing the necessary centripetal force at the top of the circle.

Question 26.
Why does a speedy motor cyclist bends towards the center of a circular path while taking a turn on it?
Answer:
So that in addition of the frictional force, the horizontal component of the normal reaction also provides the necessary centripetal forces.

Question 27.
An impulse is applied to a moving object with a force at an angle of 20° wr.t. velocity vector, what is the angle between the impulse vector and change in momentum vector ?
Answer:
Impulse and change in momentum are along the same direction. Therefore angle between these two vectors is zero.

Samacheer Kalvi 11th Physics Laws of Motion Short Answer Questions (2 Marks)

Question 28.
A man getting out of a moving bus runs in the same direction for a certain distance. Comment.
Answer:
Due to inertia of motion.

Question 29.
If the net force acting upon the particle is zero, show that its linear momentum remains constant.
Answer:
As F x \(\frac {dp}{dt}\)
when F = 0, \(\frac {dp}{dt}\) = 0 so P = constant

Question 30.
A force of 36 dynes is inclined to the horizontal at an angle of 60°. Find the acceleration in a mass of 18 g that moves in a horizontal direction.
Answer:
F = 36 dyne at an angle of 60°
Fx = F cos 60° = 18 dyne
Fx = max
So ax = \(\frac{F_{x}}{m}\) = 1 cm /s2

Question 31.
The motion of a particle of mass m is described by h = ut + \(\frac {1}{2}\) gt2. Find the force acting on particle.
Answer:
a = ut + \(\frac {1}{2}\) gt2
find a by differentiating h twice w.r.t.
a = g
As F = ma so F = mg (answer)

Question 32.
A particle of mass 0.3 kg is subjected to a force of F = -kx with k= 15 Nmr-1. What will be its initial acceleration if it is released from a point 20 cm away from the origin?
Answer:
As F = ma so F = -kx = ma
a = \(\frac {-kx}{m}\)
for x = 20 cm, ⇒ a = -10 m/s2.

Question 33.
A 50 g bullet is fired from a 10 kg gun with a speed of 500 ms-1. What is the speed of the recoil of the gun?
Answer:
Initial momentum = 0
Using conservation of linear momentum mv + MV = 0
V = \(\frac {-mv}{M}\) ⇒ V = 2.5 m/s

Question 34.
Smooth block is released at rest on a 45° incline and then slides a distance d. If the time taken of slide on rough incline is n times as large as that to slide than on a smooth incline. Show that coefficient of friction, µ = \(\left(1-\frac{1}{n^{2}}\right)\)
Answer:
When there is no friction, the block slides down the inclined plane with acceleration. a = g sin θ
when there is friction, the downward acceleration of the block is a’ = g (sin θ – µ cos θ)
As the block Slides a distance d in each case so
d = \(\frac {1}{2}\) at2 = \(\frac {1}{2}\) a’t’2
\(\frac{a}{a^{\prime}}=\frac{t^{\prime 2}}{t^{2}}=\frac{(n t)^{2}}{t^{2}}\) = n2
or \(\frac {g sin θ}{g(sin θ – µ cos θ)}\) = n2
Solving, we get (Using θ = 45°)
µ = 1 – \(\frac{1}{n^{2}}\)

Question 35.
A spring balance is attached to the ceiling of a lift. When the lift is at rest spring balance reads 49 N of a body hang on it. If the lift moves:

  1. Downward
  2. upward, with an acceleration of 5 ms2
  3. with a constant velocity.

What will be the reading of the balance in each case?
Answer:
1.  R = m(g – a) = 49 N
so = m = \(\frac {49}{9.8}\) = 5 kg
R = 5 (9.8 – 5)
R = 24 N

2. R = m(g + a)
R = 5 (9.8 + 5)
R = 74 N

3.  as a = 0 so R = mg = 49 N

Question 36.
A bob of mass 0.1 kg hung from the ceiling of room by a string 2 m long is oscillating. At its mean position the speed of a bob is 1 ms-1. What is the trajectory of the ‘oscillating bob if the string is cut when the bob is –

  1. At the mean position
  2. At its extreme position.

Answer:

  1. Parabolic
  2. vertically downwards

Question 37.
A block placed on a rough horizontal surface is pulled by a horizontal force F. Let f be the force applied by the rough surface on the block. Plot a graph of f versus F.
Answer:

Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Unto point A, f = F (50 Long as block is stationary) beyond A, when F increases, block starts moving f remains constant.

Question 38.
A mass of 2 kg is suspended with thread AB. Thread CD of the same type is attached to the other end of 2 kg mass.

  • Lower end of the lower thread is pulled gradually, hander and hander is the downward direction so as to apply force on AB Which of the thread will break & why?
  • If the lower thread is pulled with a jerk, what happens?

Answer:

  • Thread AB breaks down
  • CD will break.

Question 39.
A block of mass M is held against a rough vertical wall by pressing it with a finger. If the coefficient of friction between the block and the wall is p and the acceleration due to gravity is g, calculate the minimum force required to be applied by the finger to held the block against the wall?
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
For the block not to fall f = Mg
But f = µR = µF so
µF = Mg
F = \(\frac {Mg}{µ}\)

Samacheer Kalvi 11th Physics Laws of Motion Short Answer Questions (3 Marks) & Numericals

Question 40.
A block of mass 500 g is at rest on a horizontal table. What steady force is required to give the block a velocity of 200 cm s-2 in 4 s?
Answer:
Use F – ma
a = \(\frac {v – u}{ t }\) = \(\frac {200- 0}{ 4 }\) = 50 cm/s2
F = 500 x 50 = 25,000 dyne.

Question 41.
A force of 98 N is just required to move a mass of 45 kg on a rough horizontal surface. Find the coefficient of friction and angle of friction?
Answer:
F = 48 N,R = 45 x 9.8 = 441 N
µ = \(\frac {F’}{ R}\) = 0.22
Angle of friction θ = tan-1 0.22 = 12°24′

Question 42.
Calculate the force required to move a train of 2000 quintal up on an incline plane of 1 in 50 with an acceleration of 2 ms-2. The force of friction per quintal is 0.5 N.
Answer:
Force of friction = 0.5 N per quintal
f = 0.5 x 2000 = 1000 N
m = 2000 quintals = 2000 x 100 kg
sin θ = \(\frac {1}{50}\), a – 2 m/s2
In moving up an inclined plane, force required against gravity
mg sin θ = 39200 N
And force required to produce acceleration = ma
= 2000 x 100 x 2 = 40,0000 N
Total force required = 1000 + 39,200 + 40,0000 = 440200 N.

Question 43.
A force of 100 N gives a mass m1, an acceleration of 10 ms-2 and of 20 ms-2 to a mass m2.
What acceleration must be given to it if both the masses are tied together?
Answer:
Suppose, a = acceleration produced if m1 and m2 are tied together,
F = 100 N
Let a1 and a2 be the acceleration produced in m1 and m2 respectively.
∴ a1 and a2 = 20ms-2 (given)
Again m1 = \(\frac{\mathrm{F}}{a_{1}}\) and m2 = \(\frac{\mathrm{F}}{a_{2}}\)
⇒ m1 = \(\frac {100}{10}\) = 10kg
and m2 = \(\frac {100}{20}\) = 5kg
∴ m1 + m2 = 10 + 5 = 15
so, a = \(\frac{\mathbf{F}}{m_{1}+m_{2}}\) = \(\frac {100}{15}\) = \(\frac {20}{3}\) = 6.67 ms2

Question 44.
The pulley arrangement of figure are identical. The mass of the rope is negligible. In (a) mass m is lifted up by attaching a mass (2m) to the other end of the rope. In (b), m is lifted up by pulling the other end of the rope with a constant downward force F = 2 mg. In which case, the acceleration of m is more?
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Answer:
Case (a):
a = \(\frac {2m – m}{2m + m}\) g = a = \(\frac {g}{3}\)
Case (b):
FBD of mass m
ma’ = T – mg
ma’ = 2 mg – mg
⇒ ma’ = mg
a’ = g
So in case (b) acceleration of m is more.

Question 45.
Figure shows the position-time graph of a particle of mass 4 kg. What is the
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
(a) Force on the particle for t < 0, t > 4s, 0 < t < 4s?
(b) Impulse at t = 0 and t = 4s?
(Consider one dimensional motion only)
Answer:
(a) For t < 0. No force as Particles is at rest. For t > 4s, No force again particle comes at rest.
For 0 < t < 4s, as slope of OA is constant so velocity constant i.e., a = 0, so force must be zero.

(b) Impulse at t = 0
Impulse = change in momentum
I = m(v – w) = 4(0 – 0.75) = 3 kg ms-1
Impulse at t = 4s
1 = m(v – u) = 4 (0 – 0.75) = -3 kg ms-1

Question 46.
What is the acceleration of the block and trolley system as the figure, if the coefficient of kinetic friction between the trolley and the surface is 0.04? Also Calculate friction in the string: Take g = 10 m/s2, mass of the string is negligible.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Answer:
Free body diagram of the block
30 – T = 3a
Free body diagram of the trolley
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
T – fk = 20 a ………….(2)
where fk = µk= 0.04 x 20 x 10 = 8 N
Solving (i) & (ii), a = 0.96 m/s2 and T = 27.2 N

Question 47.
Three blocks of masses ml = 10 kg, m2 = 20 kg are connected by strings on smooth horizontal surface and pulled by a force of 60 N. Find the acceleration of the system and frictions in the string.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Solution:
All the blocks more with common acceleration a under the force F = 60 N.
F = (m1 + m2 + m3)a
a = \(\frac{\mathrm{F}}{\left(m_{1}+m_{2}+m_{3}\right)}\) = 1 m/s2
to determine, T1 →Free body diagram of m1.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
T1 = m1a = 10 x 1 = 10 N
to determine, T2 →Free body diagram of m3
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
F – T2 = m3a
Solving, we get T2 = 30 N

Question 48.
The rear side of a truck is open and a box of 40 kg mass is placed 5m away from the open end. The coefficient of friction between the box and the surface below it is 0.15 on a straight road, the truck starts from rest and accelerates with 2 m/s2. At what distance from the starting point does the box fall off the truck ? (ignore the size of the box)
Answer:
Force on the box due to accelerated motion of the truck
F = ma = 40 x 2 = 80 N (in forward direction)
Reaction on the box, F’ = F = 80 N (in backward direction)
Force of limiting friction, f = µR = 0 .15 x 40 x 10 = 60 N
Net force on the box in backward direction is P = F’ f = 80 – 60 = 20 N
Backward acceleration in the box = a= \(\frac {p}{m}\) = \(\frac {20}{40}\) = 0.5 ms-2
t = time taken by the box to travel s = 5 m and falls off the truck, then from
s = ut + \(\frac {1}{2}\) at2
5 = 0 x t + \(\frac {1}{2}\) x 0.5 x t2
t = 4.47
If the truck travels a distance x during this time
then x = 0 x 4.34 +\(\frac {1}{2}\) x 2 x (4.471)2
x = 19.98 m

Question 49.
A block slides down as incline of 30° with the horizontal. Starting from rest, it covers 8 m in the first 2 seconds. Find the coefficient of static friction.
Use s = ut + \(\frac {1}{2}\) at2
a = \(\frac{2 s}{t^{2}}\) at2 as u = 0
µ = \(\frac{g sin θ – a}{g Cos θ }\)
Putting the value and solving, µ = 0.11

Question 50.
A helicopter of mass 2000 kg rises with a vertical acceleration of 15 m/s2 . The total mass of the crew and passengers is 500 kg. Give the magnitude and direction of the:
(a) Force on the floor of the helicopter by the crew and passenger.
(b) Action of the rotor of the helicopter on the surrounding air
(c) Force on the helicopter due to the surrounding air (g = 10 m/s2 )
Answer:
(a) Force on the floor of the helicopter by the crew and passengers
= apparent weight of crew and passengers
= 500(10+ 15)
=12500 N

(b) Action of rotor of helicopter on surrounding air is Obviously vertically downwards, because helicopter rises on account of reaction of this force. Thus force of action
= (2000 + 500) (10 + 15)
= 2500 x 25
= 62,500 N

(c) Force on the helicopter due to surrounding air is obviously a reaction. As action and reaction are equal and opposite, therefore
Force of reaction F’ = 62,500 vertically upwards.

Question 51.
A rectangular box lies on a rough inclined surface. The coefficient of friction between the surface and the box is (µ). Let the mass of the box be m.

  1.  At what angle of inclination θ of the plane to the horizontal will the box just start to slide down the plane ?
  2. What is the force acting on the box down the plane, if the angle of inclination of the plane is increased to a > θ.
  3. What is the force needed to be applied upwards along the plane to make the box either remain stationary or just move up with uniform speed ?
  4. What is the force needed to be applied upwards along the plane to pk kg f make the box move up the plane with acceleration a ?

Answer:
1. When the box just starts sliding
µ = tanθ
or 0 = tan-1 µ

2. Force acting on the box down the plane
= mg (sin a – µ cos a)

3. Force needed mg (sin a + µ cos a)

4. Force needed = mg (sin a + µ cos a) + ma.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Question 52.
Two masses of 5 kg and 3 kg are suspended with help of mass less in extensible string as shown. Calculate T1 and T2 when system is going upwards with acceleration m/s2. (Use g 9.8 m/s2)
Answer:
According Newton’s second law of motion
(1) T1 – (m1 + m2)g = (m1 + m2)a
T1 = (m1 + m2)(a + g) = (5 + 3) (2 + 9.8)
T1 = 94.4 N

(2) T2 – m2g = m2a
T2 = m2 (a + g)
T2 = 3(2 + 9.8)
T2 = 35.4 N

Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Question 53.
There are few forces acting at a Point P produced by strings as shown, which is at rest. Find the forces F1 & F1
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Answer:
Using Resolution of forces IN and 2N and then applying laws of vector addition. Calculate for F1 & F1.
F1 = \(\frac{1}{\sqrt{2}}\) N, F2 = \(\frac{3}{\sqrt{2}}\)N

Question 54.
A hunter has a machine gun that can fire 50g bullets with a velocity of 150 ms A 60 kg tiger springs at him with a velocity of 10 ms-1. How many bullets must the hunter fire into the target so as to stop him in his track?
Answer:
Given m = mass of bullet = 50 gm = 0.50 kg
M = mass of tiger = 60 kg
v = Velocity of bullet – 150 m/s
V = Velocity of tiger = – 10 m/s
(v It is coming from opposite direction n = no. of bullets fired per second at the tiger so as to stop it.)
Pi = 0, before firing ……..(i)
Pf = n (mv) + MV …………(ii)
∴ From the law of conservation of momentum,
Pi = Pf
⇒ 0 = n (mv) + MV
n = \(\frac{MV}{mv}\) = \(\frac{-60 \times(-10)}{0.05 \times 150}\) = 80

Question 55.
Two blocks of mass 2 kg and 5 kg are connected by an ideal string passing over a pulley. The block of mass 2 kg is free to slide on a surface inclined at an angle of 30° with the horizontal whereas 5 kg block hangs freely. Find the acceleration of the system and the tension in the string.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Let a be the acceleration of the system and T be the Tension in the string. Equations of motions for 5 kg and 2 kg blocks are
5g – T = 5a
T – 2g sin θ – f = 2a
where f = force of limiting friction
= µR = µ mg cos θ = 0.3 x 2 g x cos 30°
Solving (1) & (2)
a = 4.87 m/s2

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