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You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.7

10th Maths Exercise 2.7 Samacheer Kalvi Question 1.
Which of the following sequences are in G.P?
(i) 3, 9, 27, 81, ……..
(ii) 4, 44, 444, 4444, ………
(iii) 0.5, 0.05, 0.005, ……..
(iv) \(\frac { 1 }{ 3 } \),\(\frac { 1 }{ 6 } \),\(\frac { 1 }{ 12 } \)………
(v) 1, -5, 25, -125, …….
(vi) 120, 60, 30, 18, …….
(vii) 16, 4, 1, \(\frac { 1 }{ 4 } \), ……
Solution:
(i) 3, 9, 27, 81
r = Common ratio




10th Maths Exercise 2.7 Question 2.
Write the first three terms of the G.P. whose first term and the common ratio are given below.
(i) a = 6, r = 3
(ii) a = \(\sqrt { 2 }\) , r = \(\sqrt { 2 }\)
(iii) a = 1000, r = \(\frac { 2 }{ 5 } \)
Solution:
(i) a = 6, r = 3
t_n = ar^n-1
t₁ = ar^1-1 = ar⁰ = a = 6
t₂ = ar^2-1 = ar¹ = 6 × 3 = 18
t₃ = ar^3-1 = ar² = 6 × 3² = 54
∴ The 3 terms are 6, 18, 54, ….

The 3 terms are 1000, 400, 160, ……………

Ex 2.7 Class 10 Samacheer Question 3.
In a G.P. 729, 243, 81,… find t₇.
Solution:
G.P = 729, 243, 81 ……
t₇ = ?

Ex 2.7 Class 10 Question 4.
Find x so that x + 6, x + 12 and x + 15 are consecutive terms of a Geometric Progression.
Answer:
\(\frac{t_{2}}{t_{1}}\) = \(\frac { x+12 }{ x+6 } \),\(\frac{t_{3}}{t_{2}}\) = \(\frac { x+15 }{ x+12 } \)
Since it is a G.P.
\(\frac { x+12 }{ x+6 } \) = \(\frac { x+15 }{ x+12 } \)
(x + 12)² = (x + 6) (x + 15)
x² + 24x + 144 = x² + 21x + 90
3x = -54 ⇒ x = \(\frac { -54 }{ 3 } \) = -18

Exercise 2.7 Class 10 Maths Question 5.
Find the number of terms in the following G.P.
(i),4, 8, 16, …, 8192
(ii) \(\frac { 1 }{ 3 } \),\(\frac { 1 }{ 9 } \),\(\frac { 1 }{ 27 } \),……\(\frac { 1 }{ 2187 } \)
Solution:
(i) 4, 8, 16, …… 8192

10th Maths 2.7 Exercise Question 6.
In a G.P. the 9^th term is 32805 and 6^th term is 1215. Find the 12th term.
Solution:
In a G.P
t_n = ar^n-1
t₉ = 32805
t₆ = 1215
t₁₂ = ?
Let
t₉ = ar⁸ = 32805 ………(1)
t₆ = ar⁵ = 1215 ………. (2)

10th Math Exercise 2.7 Solution Question 7.
Find the 10th term of a G.P. whose 8^th term is 768 and the common ratio is 2.
Answer:
Here r = 2, t₈ = 768
t₈ = 768 (t_n = ar^n-1)
a. r^8-1 = 768
ar⁷ = 768 …..(1)
10^th term of a G.P. = a.r 10^-1
= ar⁹
= (ar⁷) × (r²)
= 768 × 2² (from 1)
= 768 × 4 = 3072
∴ 10^th term of a G.P. = 3072

Class 10 Maths Exercise 2.7 Solutions Question 8.
If a, b, c are in A.P. then show that 3^a,3^b,3^c are in G.P.
Solution:
If a, b, c are in A.P
t₂ – t₁ = t₃ – t₂
b – a = c – b
2b = c + a
To prove that 3^a, 3^b, 3^c are in G.P
⇒ 3^2b = 3^c+a + a [Raising the power both sides]
⇒ 3^b.3^b = 3^c.3^a
\(\Rightarrow \frac{3^{b}}{3^{a}}=\frac{3^{c}}{3^{b}}\)
\(\Rightarrow \frac{t_{2}}{t_{1}}=\frac{t_{3}}{t_{1}}\)
⇒ Common ratio is same for 3^a,3^b,3^c
⇒ 3^a, 3^b, 3^c forms a G.P
∴ Hence it is proved .

Exercise 2.7 Class 10 Maths Solution Question 9.
In a G.P. the product of three consecutive terms is 27 and the sum of the product of two terms taken at a time is \(\frac { 57 }{ 2 } \). Find the three terms.
Solution:
Let the three consecutive terms in a G.P are \(\frac { a }{ r } \), a, ar.
Their Product = \(\frac { a }{ r } \) × a × ar = 27
a³ = 27 = 3³
a = 3
Sum of the product of terms taken two at a time is \(\frac { 57 }{ 2 } \)

2.7 Exercise Class 10 Question 10.
A man joined a company as Assistant Manager. The company gave him a starting salary of ₹60,000 and agreed to increase his salary 5% annually. What will be his salary after 5 years?
Solution:
Starting salary = ₹ 60,000
Increase per year = 5%
∴ At the end of 1 year the increase
= 60,0,00 × \(\frac { 5 }{ 100 } \)
₹ 3000
∴ At the end of first year his salary
= ₹ 60,000 + 3000
I year salary = ₹ 63,000
II Year increase = 63000 × \(\frac { 5 }{ 100 } \)
At the end of II year, salary
= 63000 + 3150
= ₹ 66150
III Year increase = 66150 × \(\frac { 5 }{ 100 } \)
= 3307.50
At the end of III year, salary = 66150 + 3307.50
= ₹ 69457.50
IV year increase = 69457.50 × \(\frac { 5 }{ 100 } \)
= ₹ 3472.87

Exercise 2.7 Class 10 Question 11.
Sivamani is attending an interview for a job and the company gave two offers to him. Offer A: ₹ 20,000 to start with followed by a guaranteed annual increase of 3% for the first 5 years.
Offer B: ₹ 22,000 to start with followed by a guaranteed annual increase of 3% for the first 5 years.
What is his salary in the 4th year with respect to the offers A and B?
Solution:
Offer A
Starting salary ₹ 20,000
Annual increase 6%

At the end of
III year ,salary = 22472 + 1348 = 23820
∴ IV year salary = ₹ 23820
Offer B
Starting salary = ₹ 22,000

Salary as per Option A = ₹ 23820
Salary as per Option B = ₹ 24040
∴ Option B is better.

10 Class Math Exercise 2.7 Solution Pdf Question 12.
If a, b, c are three consecutive terms of an A.P. and x, y, z are three consecutive terms of a G.P. then prove that x^b-c × y^c-a × z^a-b = 1.
Solution:
a, b, c are three consecutive terms of an AP.
∴ Let a, b, c be a, a + d, a + 2d respectively ………… (1)
x, y, z are three consecutive terms of a GP.
∴ Assume x, y, z as x, x.r, x.r² respectively ……… (2)
PT : x^b-c , y^c-a , z^a-b = 1
Substituting (1) and (2) in LHS, we get
LHS = x^a+d-a-2d × (xr)^a+2d-a × (xr²)^a-a-d
= (x)^-d . (xr)^2d (xr²)^-d
= \(\frac{1}{x^{d}}\) × x^2d . r^2d × \(\frac{1}{x^{d} r^{2 d}}\) = 1 = RHS

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.2

9th Maths Exercise 1.2 Samacheer Kalvi Question 1.
Find the cardinal number of the following sets.
(i) M = {p, q, r, s, t, u}
(ii) P = {x : x = 3n + 2, n ∈ W and x < 15}
(iii) Q = {v : v = \(\frac { 4 }{ 3n }\) ,n ∈ N and 2 < n ≤ 5}
(iv) R = {x : x is an integers, x ∈ Z and -5 ≤ x < 5}
(v) S = The set of all leap years between 1882 and 1906.
Solution:
(i) n(M) = 6
(ii) W = {0, 1, 2, 3, ……. }
if n = 0, x = 3(0) + 2 = 2
if n = 1, x = 3(1) + 2 = 5
if n = 2, x = 3(2) + 2 = 8
if n = 3, x = 3(3)+ 2 =11
if n = 4, x = 3(4) + 2=14
∴ P= {2, 5, 8, 11, 14}
n(P) = 5

(iii) N = {1,2, 3, 4, …..}
n ∈ {3, 4, 5}

n(Q) = 3

(iv) x ∈ z
R = {-5, – 4, -3, -2, -1, 0, 1, 2, 3, 4}
n(R)= 10.

(v) S = {1884, 1888, 1892, 1896, 1904}
n (S) = 5.

9th Maths Set Language Exercise 1.2 Question 2.
Identify the following sets as finite or infinite.
(i) X = The set of all districts in Tamilnadu.
(ii) Y = The set of all straight lines passing through a point.
(iii) A = {x : x ∈ Z and x < 5}
(iv) B = {x : x² – 5x + 6 = 0, x ∈ N}
Solution:
(i) Finite set
(ii) Infinite set
(iii) A = { ……. , -2, -1, 0, 1, 2, 3, 4}
∴ Infinite set

(iv) x² – 5x + 6 = 0
(x – 3) (x – 2) = 0
B = {3, 2}
∴ Finite set.

9th Maths Exercise 1.2 Question 3.
Which of the following sets are equivalent or unequal or equal sets?
(i) A = The set of vowels in the English alphabets.
B = The set of all letters in the word “VOWEL”
(ii) C = {2, 3, 4, 5}
D = {x : x ∈ W, 1 < x < 5}
(iii) X = A = { x : x is a letter in the word “LIFE”}
Y = {F, I, L, E}
(iv) G = {x : x is a prime number and 3 < x < 23}
H = {x : x is a divisor of 18}
Solution:
(i) A = {a, e, i, o, u}
B = {V, O,W, E, L}
The sets A and B contain the same number of elements.
∴ Equivalent sets

(ii) C ={2, 3, 4, 5}
D = {2, 3, 4}
∴ Unequal sets

(iii) X = {L, I, F, E}
Y = {F, I, L, E}
The sets X and Y contain the exactly the same elements.
∴ Equal sets.

(iv) G = {5, 7, 11, 13, 17, 19}
H = {1, 2, 3, 6, 9, 18}
∴ Equivalent sets.

9th Maths Set Language Exercise 1.2 Solutions Question 4.
Identify the following sets as null set or singleton set.
(i) A = (x : x ∈ N, 1 < x < 2}
(ii) B = The set of all even natural numbers which are not divisible by 2.
(iii) C = {0}
(iv) D = The set of all triangles having four sides.
Solution:
(i) A = { } ∵ There is no element in between 1 and 2 in Natural numbers.
∴ Null set

(ii) B = { } ∵ All even natural numbers are divisible by 2.
∴ B is Null set

(iii) C = {0}
∴ Singleton set

(iv) D = { }
∵ No triangle has four sides.
∴ D is a Null set.

9th Maths Exercise 1.2 In Tamil Question 5.
State which pairs of sets are disjoint or overlapping?
(i) A = {f, i, a, s} and B = {a, n, f, h, s)
(ii) C = {x : x is a prime number, x > 2} and D = {x : x is an even prime number}
(iii) E = {x: x is a factor of 24} and F = {x : x is a multiple of 3, x < 30}
Solution:
(i) A = {f, i, a, s}
B = {a, n, f, h, s}
A ∩ B ={f, i, a, s} ∩ {a, n,f h, s} = {f, a, s}
Since A ∩ B ≠ ϕ , A and B are overlapping sets.

(ii) C = {3, 5, 7, 11, ……}
D = {2}
C ∩ D = {3, 5, 7, 11, …… } ∩ {2} = { }
Since C ∩ D = Ø, C and D are disjoint sets.

(iii) E = {1, 2, 3, 4, 6, 8, 12, 24}
F = {3, 6, 9, 12, 15, 18, 21, 24, 27}
E ∩ F = {1, 2, 3, 4, 6, 8, 12, 24} ∩ {3, 6, 9, 12, 15, 18, 21, 24, 27}
= {3, 6, 12, 24}
Since E ∩ F ≠ ϕ, E and F are overlapping sets.

9th Standard Maths Exercise 1.2 In Tamil Question 6.
If S = {square,rectangle,circle,rhombus,triangle}, list the elements of the following subset of S.
(i) The set of shapes which have 4 equal sides.
(ii) The set of shapes which have radius.
(iii) The set of shapes in which the sum of all interior angles is 180°
(iv) The set of shapes which have 5 sides.
Solution:
(i) {Square, Rhombus}
(ii) {Circle}
(iii) {Triangle}
(iv) Null set.

9th Standard Maths Exercise 1.2 Question 7.
If A = {a, {a, b}}, write all the subsets of A.
Solution:
A= {a, {a, b}} subsets of A are { } {a}, {a, b}, {a, {a, b}}.

9th Std Maths Exercise 1.2 Question 8.
Write down the power set of the following sets.
(i) A = {a, b}
(ii) B = {1, 2, 3}
(iii) D = {p, q, r, s}
(iv) E = Ø
Solution:
(i) The subsets of A are Ø, {a}, {b}, {a, b}
The power set of A
P(A ) = {Ø, {a}, {b}, {a,b}}

(ii) The subsets of B are ϕ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}
The power set of B
P(B) = {Ø, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}}

(iii) The subset of D are Ø, {p}, {q}, {r}, {s}, {p, q}, {p, r}, {p, s}, {q, r}, {q, s}, {r, s},{p, q, r}, {q, r, s}, {p, r, s}, {p, q, s}, {p, q, r, s}}
The power set of D
P(D) = {Ø, {p}, {q}, {r}, {s}, {p, q}, {p, r}, {p, s}, {q, r}, {q, s}, {r, s}, {p, q, r}, {q, r, s}, {p, r, s}, {p, q, s}, {p, q, r, s}

(iv) The power set of E
P(E) = { }.

9th Maths 1.2 In Tamil Question 9.
Find the number of subsets and the number of proper subsets of the following sets.
(i) W = {red,blue, yellow}
(ii) X = { x² : x ∈ N, x² ≤ 100}.
Solution:
(i) Given W = {red, blue, yellow}
Then n(W) = 3
The number of subsets = n[P(W)] = 2³ = 8
The number of proper subsets = n[P(W)] – 1 = 2³ – 1 = 8 – 1 = 7

(ii) Given X ={1,2,3, }
X² = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}
n(X) = 10
The Number of subsets = n[P(X)] = 2¹⁰ = 1024
The Number of proper subsets = n[P(X)] – 1 = 2¹⁰ – 1 = 1024 – 1 = 1023.

9th Maths 1.2 Exercise Question 10.
(i) If n(A) = 4, find n[P(A)].
(ii) If n(A) = 0, find n[P(A)].
(iii) If n[P(A)] = 256, find n(A).
Solution:
(i) n( A) = 4
n[ P(A)] = 2n = 2⁴ = 16
(ii) n(A) = 0
n[P(A)] = 2⁰ = 1
(iii) n[P(A)] = 256

n[P(A)] = 28
∴ n(A) = 8.

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Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Students looking for Chapter 6 Gaseous State Concepts can find them all in one place from our Tamilnadu State Board Solutions. Simply click on the links available to prepare the corresponding topics of Chemistry easily. Samacheer Kalvi 11th Chemistry Chapter wise Questions and Answers are given to you after sample research and as per the latest edition textbooks. Clarify all your queries and solve different questions to be familiar with the kind of questions appearing in the exam. Thus, you can increase your speed and accuracy in the final exam.

Samacheer Kalvi 11th Chemistry Gaseous State Textual Evaluation Solved

I. Choose the correct answer from the following:

11th Chemistry Lesson 6 Book Back Answers Question 1.
Gases deviate from ideal behavior at high pressure. Which of the following statement (s) is correct for non – ideality?
(a) at high pressure the collision between the gas molecule become enormous
(b) at high pressure the gas molecules move only in one direction
(c) at high pressure, the volume of gas become insignificant
(d) at high pressure the inter molecular interactions become significant
Answer:
(d) at high pressure the inter molecular interactions become significant

11th Chemistry Chapter 6 Book Back Answers Question 2.
Rate of diffusion of a gas is …………
(a) directly proportional to its density
(b) directly proportional to its molecular weight
(c) directly proportional to its square root of its molecular weight
(d) inversely proportional to the square root of its molecular weight
Answer:
(d) inversely proportional to the square root of its molecular weight

Gaseous State 11th Chemistry Question 3.
Which of the following is the correct expression for the equation of state of van der Waals gas?

Answer:

11th Chemistry Unit 6 Book Back Answers Question 4.
When an ideal gas undergoes unrestrained expansion, no cooling occurs because the molecules ………….
(a) are above inversion temperature
(b) exert no attractive forces on each other
(c) do work equal to the loss in kinetic energy
(d) colide without loss of energy
Answer:
(b) exert no attractive forces on each other

11th Chemistry Gaseous State Question 5.
Equal weights of methane and oxygen is mixed in an empty container at 298 K. The fraction of total pressure exerted by oxygen ………..
(a) 1/3
(b) 1/2
(c) 2/3
(d) 1/3 × 273 × 298
Answer:
(a) 1/3
Hint:
mass of methane = mass of oxygen = a
number of moles of methane = \(\frac {a}{16}\)
number of moles of Oxygen = \(\frac {a}{32}\)
mole fraction of Oxygen =
Partial pressure of oxygen = mole fraction x Total Pressure = \(\frac {1}{3}\)P

Gaseous State Class 11 Question 6.
The temperatures at which real gases obey the ideal gas laws over a wide range of pressure is called …………
(a) Critical temperature
(b) Boyle temperature
(c) Inversion temperature
(d) Reduced temperature
Answer:
(b) Boyle temperature
Hint:
The temperature at which real gases obey the ideal gas laws over a wide range of pressure is called Boyle temperature

Gaseous State Class 11 Notes Pdf Question 7.
In a closed room of 1000 m³ a perfume bottle is opened up. The room develops a smell. This is due to which property of gases?
(a) Viscosity
(b) Density
(c) Diffusion
(d) None
Answer:
(c) Diffusion

Samacheer Kalvi Guru 11th Chemistry Question 8.
A bottle of ammonia and a bottle of HCl connected through a long tube are opened simultaneously at both ends. The white ammonium chloride ring first formed will be ………….
(a) At the center of the tube
(b) Near the hydrogen chloride bottle
(c) Near the ammonia bottle
(d) Throughout the length of the tube
Answer:
(b) Near the hydrogen chloride bottle
Hint:
Rate of diffusion α 1/√m
m_NH3  = 17
m_HCl = 36.5
γ_NH3  > γ_HCl
Hence white fumes first formed near hydrogen chloride.

Samacheer Kalvi Class 11 Chemistry Solutions Question 9.
The value of universal gas constant depends upon ………..
(a) Temperature of the gas
(b) Volume of the gas
(c) Number of moles of the gas
(d) units of Pressure and volume
Answer:
(d) units of Pressure and volume

Samacheer Kalvi.Guru 11th Chemistry Question 10.
The value of the gas constant R is …………
(a) 0.082 dm³ atm.
(b) 0.987 cal mol^-1 K^-1
(c) 8.3 J mol^-1K^-1
(d) 8 erg mol^-1K^-1
Answer:
(c) 8.3 J mol^-1K^-1

Samacheerkalvi.Guru 11th Chemistry Question 11.
Use of hot air balloon in sports at meteorological observation is an application of
(a) Boyle’s law
(b) Newton’s law
(c) Kelvin’s law
(d) Brown’s law
Answer:
(a) Boyle’s law

Class 11 Chemistry Solutions Samacheer Kalvi Question 12.
The table indicates the value of van der Waals constant ‘a’ in (dm³)² atm. mol^-2

The gas which can be most easily liquefied is ………….
(a) O₂
(b) N₂
(c) NH₃
(d) CH₄
Answer:
(c) NH₃
Hint:
Higher the value of ‘a’, greater the intermolecular force of attraction, easier the liquefaction. Option (c) is correct

Gaseous State Questions And Answers Pdf Question 13.
Consider the following statements.
(i) Atmospheric pressure is less at the top of a mountain than at sea level
(ii) Gases are much more compressible than solids or liquids
(iii) When the atmospheric pressure increases the height of the mercury column rises Select the correct statement.
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i) and (iii)
(d) (i), (ii) and (iii)
Answer:
(d) (i), (ii) and (iii)

Class 11 Gaseous State Question 14.
Compressibility factor for CO₂ at 400 K and 71.0 bar is 0.8697. The molar volume of CO₂ under these conditions is ………..
(a) 22.04 dm³
(b) 2.24 dm³
(c) 0.41 dm³
(d) 19.5 dm³
Answer:
(c) 0.41 dm³
Compressibility factor (z) = \(\frac {Pv}{nRT}\)
V = \(\frac {z x nRT }{p}\)

V = 0.41 dm³

Samacheer Kalvi 11th Chemistry Solution Question 15.
If temperature and volume of an ideal gas is increased to twice its values, the initial pressure P becomes ………….
(a) 4P
(b) 2P
(c) P
(d) 3P
Answer:
(c) P
Hint:

P₂ = P₂ Option (c)

Question 16.
At identical temperature and pressure, the rate of diffusion of hydrogen gas is 3\(\sqrt{3}\) times that of a hydrocarbon having molecular formula What is the value of n?
(a) 8
(b) 4
(c) 3
(d) 1
Answer:
(b) 4.
Hint:

Squaring on both sides and rearranging
27 x 2 = m_CnH_2n-2
54 = n(12) + (2n-2)(l)
54 = 12n+2n – 2
54 = 14n – 2
n = (54 + 2)/14 = 56/14 = 4

Question 17.
Equal moles of hydrogen and oxygen gases are placed in a container, with a pin-hole through which both can escape what fraction of oxygen escapes in the time required for one-half of the hydrogen to escape. (NEET phase 1)
(a) 3/8
(b) 1/2
(c) 1/8
(d) 1/4
Answer:
(c) 1/8
Hint:

The fraction of oxygen that escapes in the time required for one half of the hydrogen to escape is 1/8

Question 18
The variation of volume V, with temperature T, keeping pressure constant is called the coefficient of thermal expansion ie α = \(\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_{P}\)For an ideal gas α is equal to ………..
(a) T
(b) 1/T
(c) P
(d) none of these
Answer:
(b) 1/T
Hint:

Question 19.
Four gases P, Q, R and S have almost same values of ‘b’ but their ‘a’ values (a, b are Van der Waals Constants) are in the order Q < R < S < P. At a particular temperature, among the four gases the most easily liquefiable one is ………….
(a) P
(b) Q
(c) R
(d) S
Answer:
(a) P
Hint:
Greater the ‘a’ value, casier the liquefaction

Question 20.
Maximum deviation from ideal gas is expected from (NEET)
(a) CH_4(g)
(b) NH_3(g)
(c) H_2(g)
(d) N_2(g)
Answer:
(b) NH_3(g)

Question 21.
The units of Van der Waals constants ‘b’ and ‘a’ respectively
(a) mol L^-1 and L atm² mol^-1
(b) mol L and L atm mol²
(c) mol^-1 L and L² atm mol^-1
(d) none of these
Answer:
(c) mol^-1 L and L² atm mol^-1
Hint:
an²/V² atm
a = atm L²/mol² = L² mol^-2 atm
nb = L
b = L /mol = L mol^-1

Question 22.
Assertion : Critical temperature of CO₂ is 304 K, it can be liquefied above 304 K.
Reason : For a given mass of gas, volume is to directly proportional to pressure at constant temperature.
(a) both assertion and reason arc true and 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:
(d) both assertion and reason are false
Hint:
Correct Statement: Critical temperature of CO₂ is 304 K. It means that CO₂ cannot be liquefied above 304 K, whatever the pressure may applied. Pressure is inversely proportional to volume.

Question 23.
What is the density of N, gas at 227°C and 5.00 atm pressure? (R = 0.082 L atm K^-1 mol^-1)
(a) 1.40 g/L
(b) 2.81 g/L
(c) 3.41 g/L
(d) 0.29 g/L
Answer:
(c) 3.41 g/L
Hint:
Density = \(\frac {Mass}{Volume}\)

Question 24.
Which of the following diagrams correctly describes the behaviour of a fixed mass of an ideal gas ? (T is measured in K)

Answer:

For a fixed mass of an ideal gas V α T
P α 1/V
and PV = Constant

Question 25.
25 g of each of the following gases are taken at 27°C and 600 mm Hg pressure. Which of these will have the least volume?
(a) HBr
(b) HCl
(c) HF
(d) HI
Answer:
(d) HI
Hint:
At a given temperature and pressure
Volume α number of moles
Volume α Mass / Molar mass
Volume α 28 / Molar mass
i.e. if molar mass is more , volume is less. Hence Hl has the least volume.

II. Answer these questions briefly.

Question 26.
State Boyle’s law.
Answer:
Boyle’s law states that at a given temperature. the volume occupied by a fixed mass of a gas is inversely proportional to its pressure.
V α \(\frac {1}{P}\);
where T and n are fixed or PV = Constant = k

Question 27.
A balloon filled with air at room temperature and cooled to a much lower temperalure can be used as a model for Charles’ law.
Answer:
Charles’ law:

• V T at constant P and n (or) \(\frac {V}{T}\) = Constant
• A balloon filled with air at room temperature and cooled to a much lower temperature. the size of the balloon is reduced. Because if the temperature of the gas decreases, the volume also decreases in a direct proportion.
• When temperature is reduced, the gas molecules inside in over slower due to decreased temperature and hence the volume decreases.

Question 28.
Name two items that can serve as a model for Gay Lussac’s law and explain.
Answer:
Gay Lussac’s law:
1. P α T at constant volume (or) = \(\frac {V}{T}\)
2. Example – 1:
You fill the car type completely full of air on the hottest day of summer. The type cannot change it shape and volume. But when winter comes, the pressure inside the lyre is reduced and the shape is also reduced. This confirms that pressure and temperature are direct related to each other.
3. Example – 2:
The egg in the bottle experiment.

A glass bottle is taken, inside the bottle put some pieces of cotton with fire. Then place a boiled egg (shell removed) at the top of the bottle. The temperature inside the bottle increases from the fire, rising (he pressure. By scaling the bottle with egg, the fire goes on, dropping the temperature and pressure. This causes the egg to be sucked into the bottle.
P α T is proved (or) = \(\frac{P_{1}}{V_{1}}=\frac{P_{2}}{V_{2}}\)

Question 29.
Give the mathematical expression that relates gas volume and moles. Describe in words what the mathematical expression means.
Answer:

1. The mathematical relationship betwêen the volume of a gas and the number of moles is V α n
2. \(\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}\) = Constant
   Where V₁ and n₁ are the volume and number of moles of a gas and V₂ and n₂ are the values of volume and number of moles of same gas at a different set of conditions.
3. If the volume of the gas increase then the number of moles of the gas also increases.
4. At a certain temperature and pressure, the volume of a gas is dirctly proportional to the number of the moles of the gas.

Question 30.
What are ideal gases? In what way real gases differ from ideal gases.
Answer:

1. Ideal gases are the gases that obey gas laws or gas equation PV = nRT.
2. Real gases do not obey gas equation. PV = nRT.
3. The deviation of real gases from ideal behaviour is measure in terms of a ratio of PV to nRT. This is termed as compression factor (Z). Z = \(\frac {PV}{nRT}\)
4. For ideal gases Z = 1.
5. For real gases Z > 1 or Z < 1. For example, at high pressure real gases have Z >1 and at intermediate pressure Z < 1.
6.  Above the Boyle point Z> 1 for real gases and below the Boyle point, the real gases first show a decrease for Z, reaches a minimum and then increases with the increase in pressure.
7. So, it is clear that at low pressure and high temperature, the real gases behave as ideal gases.

Question 31.
Can a Van der Waals gas with a = 0 be liquefied? Explain.
Answer:

• a = 0 for a Van der Waals gas i.e. for a real gas. Van der Waals constant a = 0. It cannot be liquefied.
• If a = 0, there is a very less interaction between the molecules of gas.
• ‘a’ is the measure of strength of Van der Waals force of attraction between the molecules of the gas.
• If a is equal to zero, the Van der Waals force of attraction is very less and the gas cannot be liquefied.

Question 32.
Suppose there ¡s a tiny sticky area on the wan of a container of gas. Molecules hitting this area stick there permanently. Is the pressure greater or less than on the ordinary area of walls?
Answer:

• Molecules hitting the tiny sticky area on the wall of the container of gas moves faster as they get closer to adhesive surface, but this effect is not permanent.
• The pressure on the sticky wall is greater than on the ordinary area of walls.

Question 33.
Explain the following observations
(a) Aerated water bottles are kept under water during summer
(b) Liquid ammonia bottle is cooled before opening the seal
(c) The type of an automobile is inflated to slightly lesser pressure in summer than in winter
(d) The size of a weather balloon becomes larger and larger as it ascends up into larger altitude
Answer:
(a) In aerated water bottles, CO₂ gas is passed through the aqueous solution under pressure because the solubility of the gas in water is not very high. In summer, the solubility of the gas in water is likely to decrease because of the rise in temperature. Thus, in summer, more of gas will be present above the liquid surface in the glass bottle.

In case, the pressure of the gas becomes too high, the glass will not be able to withstand the pressure and the bottle may explode. To avoid this, the bottles are kept under water. As a result, the temperature is likely to decrease and the solubility of CO₂ is likely to increase in aqueous solution resulting in decreased pressure.

(b) Liquid ammonia bottle contains the gas under very high pressure. If the bottle is opened as such, then the sudden decrease in pressure will lead to a large increase in volume of the gas. As a result, the gas will come out of the bottle all of a sudden with force. This will lead to the breakage of the bottle and also causes accident.

However, if the bottle is cooled under tap water for sometime, there will be a decrease in the volume of a gas to a large extent. if the seal is opened now, the gas will corne out of the bottle at a slower rate, reduces the chances of accident.

(c) The pressure of air is directly proportional to the temperature. Since the temperature is higher in summer than in higher, the pressure of the air in the tube of the lyre is likely to be quite high as compared to winter. It is quite likely that the tube may burst under high pressure in summer, Therefore, it is advisable to inflate the types to lesser pressure in summer than in winter.

(d) The volume of the gas is inversely proportional to pressure at a given temperature according to Boyle’s law. As the weather balloon ascends, the pressure tends to decrease. As a result, the volume of the gas inside the balloon or the size of the balloon is likely to increase.

Question 34.
Give suitable explanation for the following facts about gases.
(a) Gases don’t settle at the bottom of a container
(b) Gases diffuse through all the space available to them and
(c) Explain with an increase in temperature
Answer:
(a) Gases by definition are the least dense state of matter. They have negligible intermolecular forces of attraction. So they are all free to roam separately. So the least dense gas particles will not sink at the bottom of a container.

(b) When a sample of a gas introduced to one part of a closed container, its molecules very quickly disperse throughout the container, this process by which molecules disperse in space in response to differences in concentration is called diffusion. For e.g., you can smell perfume in a room, because it difluses into the air totally inside the room.

(c) Diffusion is faster at higher temperature because the gas molecules have greater kinetic energy. Since heat increase the motion, then diffusion happens faster.

Question 35.
Suggest why there ¡s no hydrogen (H₂) in our atmosphere. Why does the moon have no atmosphere?
Answer:
1. Hydrogen is the lightest element thus when produced in free state, it rises above all the other gases to the top of the atmosphere, where it is open to cosmic storms and solar flares. There it literally leaks from the atmosphere to the empty space. Hydrogen easily gains velocity required to escape Earth’s magnetic field. Hydrogen is very reactive in nature. So it would have reacted with O₂ , in its way to produce H₂O. So majority portion of H₂ reacts and very less amount of it present in the upper level of atmosphere and gains velocity to escape the atmosphere.

2. Moon has no atmosphere because the value of acceleration due to gravity ‘g’ on the surface of the moon is small. Therefore, the value of escape velocity on the surface of the moon is very small. The molecule of the atmospheric gases on the surface of the moon have thermal velocities greater than the escape velocity. That’s why all the molecules of gases have escaped and there is no atmosphere in the moon. The moon has insufficient gravity to retain an atmosphere. So we conclude that moon has no atmosphere.

Question 36.
Explain whether a gas approaches ideal behaviour or deviates from ideal behaviour if –
(a) it is compressed to a smaller volume at constant temperature
(b) the temperature is raised while keeping the volume constant
(c) more gas is introduced into the same volume and at the same temperature
Answer:
(a) it a gas is compressed to a smaller volume at constant temperature, pressure is increased. At high pressure with a smaller volume, the gas deviates from ideal behaviour.

(b) If a gas temperature is raised keeping the volume constant, the pressure of the gas will increase. At high pressure, the gas deviates from ideal behaviour.

(c) if more gas is introduced into the same volume and at the same temperature, the number of moles are increasing. if the volume remains same, the increased number of moles collide with each other and kinetic energy increases and pressure decreases. At increased pressure, the gas deviates from ideal behaviour.

Question 37.
Which of the following gases would you expect to deviate from ¡deal behaviour under conditions of low temperature F Cl₂, or Br₂? Explain.
Answer:
1. Bromine deviates (Br₂) from the ideal gas maximum than Cl₂ and F₂. Because Br₂ has biggest size (atomic weight 79.9) provides maximum attraction between bromine molecules which is directly proportional to the size of the molecule and the boiling point of the liquid made from those molecules.

2. Br₂ deviates from ideal behaviour because it has largest atomic radii compared to Cl₂ and F₂. So it contains more electrons than other two, and the Vander Waals forces are stronger in Br₂ than in Cl₂ and F₂. So Br₂ deviates from ideal behaviour.

Question 38.
Distinguish between diffusion and effusion.
Answer:
Diffusion:

• Diffusion is the spreading of molecules of a substance throughout a space or a second substance.
• Diffusion refers to the ability of the gases to mix with each other.
• E.g.. Spreading of something such as brown tea liquid spreading through the water in a tea cup.

Effusion:

• Effusion is the escape of gas molecules through a very small hole in a membrane into an evacuated area.
• Effusion is a ability of a gas to travel through a small pin-hole.
• E.g., pouring out something like the soap studs bubbling out from a bucket of water.

Question 39.
Aerosol cans carry clear warning of heating of the can. Why?
Answer:
Aerosol cans carry clear warning of heating of the can. As the temperature rises, pressure in the can will increase and ambient temperatures about 120°F may lead to explosions. So aerosol cans should always be stored in dry areas where they will not be exposed to excessive temperatures. You should never throw an aerosol can onto a fire or leave it in the direct sunlight. even it is empty. This is because the pressure will build up so much that the can will burst. It is due to 2 reasons.

• The gas pressure increases.
• More of the liquefied propellant turns into a gas.

Question 40.
When the driver of an automobile applies brake, the passengers are pushed toward the front of the car but a helium balloon is pushed toward back of the car. Upon forward acceleration the passengers are pushed toward the front of the car. Why?
Answer:
1. When the driver of an automobile applies brake, the passengers are pushed toward the front of the car due to inertia of the body, but a helium balloon pushed toward back of the car. Helium balloon responds to the air around it. Helium molecules are lighter than air of our atmosphere, and so they move toward back by gravity as a result of the accelerating frame.

2. Upon forward acceleration, the passengers arc pushed toward the front of the car, because the body in motion tends to stay in motion until acted upon by an outside force. Helium balloon is going to move opposite to this pseudo gravitational force.

Question 41.
Would it be easier to drink water with a straw on the top of Mount Everest?
Answer:
It is difficult to drink water with a straw on the top of Mount Everest. This is because the reduced atmospheric pressure is less effective in pushing water into the straw at the top of the mountain because gravity falls off gradually with height. The air pressure falls off, there isn’t enough atmospheric pressure to push the water up in the straw all the way to the mouth.

Question 42.
Write the Van der Waals equation for a real gas. Explain the correction term for pressure and volume.
Answer:
Van der Waals equation of state for real gases is –
\(\left(P+\frac{a n^{2}}{V^{2}}\right)(V-n b)\) = nRT

Correction term for pressure:
\(\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}\) is the pressure correction. It represents the intermolecular interaction that causes the non ideal behaviour.

Correction term for Volume:
V – nb is the volume correction. it is the effective volume occupied by real gas.

Question 43.
Derive the values of critical constants from the Van der Waals constants.
Answer:
Derivation of critical constants from the Van der Waals constants:
Van der Waals equation is,
\(\left(P+\frac{a n^{2}}{V^{2}}\right)(V-n b)\) = nRT for 1 mole
From this equation, the values of critical constant P_CV_C and T_C arc derived in terms of a and b the Van der Waals constants.
\(\left(P+\frac{a n^{2}}{V^{2}}\right)(V – b)\) = RT ………..(1)
On expanding the equestion (1)
P V + \(\frac {a}{V}\) – pb – \(\frac{\mathrm{ab}}{\mathrm{V}^{2}}\) – RT = 0 ………(2)
Multiplying eqestion (2) by \(\frac{V^{2}}{P}\),

equation (3) is rearranged in the powers of V
V³ – \(\left[\frac{\mathrm{RT}}{\mathrm{P}}+\mathrm{b}\right]\) V² + \(\frac {aV}{P}\) –
= 0 ………..(4)
The above equation (4) is an cubic equation of V, which can have three roots. At the critical point. all the three values of V are equal to the critical volume V_C.
i.e. V = V_C.
V – V_C = O ……….(5)
(V – V_C)³ = O ………(6)
(V³ – 3V_CV² + 3V_C³V – V_C³ = 0 ………(7)
As the equation (4) is identical with equation (7), comparing the ‘V’ ternis in (4) and (7),

Divide equation (11) by (10)

When equation (12) is substituted in (10)

substituting the values of Vc and Pc in equation (9)

Critical constant a and b can be calculated using Van der Waals Constant as follows:

Question 44.
Why do astronauts have to wear protective suits when they are on the surface of moon?
Answer:
In space, there is no pressure, if we do wear a pressurised suit, our body will die. In space, we have to wear a pressurised suit, otherwise our body will continue to push out and blow up like a balloon. It would look cool, but we will be dead. So the astronauts in space must wear a pressurised suit (protective suits).

Question 45.
When ammonia combines with HCl, NH₄ Cl is formed as white dense fumes. Why do more funies appear near HCl?
Answer:

1. When ammonia combines with HCl, NH₄ Cl is formed as white dense fumes. The reaction takes place in neutralization between a weak base and a strong acid.
2. The property of the gas is diffusion.
3. Diffusion of gases Ammonia and hydrogen chloride. Concentrated ammonia solution is placed on a pad in one end of a tube and concentrated HCl on the pad at the other. After about a minute, the gases diffuses far enough to meet and a ring of solid ammonium chloride is formed near the HCl end.

Question 46.
A sample of gas at 15°C at 1 atm has a volume of 2.58 dm³. Vhen the temperature is raised to 38°C at I atm does the volume of the gas increase? if so, calculate the final volume.
Answer:

V₂ = 2.78 dm³ i.e. volume increased from 2.58 dm³ to 2.78 dm³.

Question 47.
A sample of gas has a volume of 8.5 dm³ at an unknown temperature. When the sample is submerged in ice water at 0°C, its volume gets reduced to 6.37 dm³. What ¡s its initial temperature?
Answer:

T₁ = 364.28 K

Question 48.
Of two samples of nitrogen gas, sample A contains 1.5 moles of nitrogen in a vessel of volume of 37.6 dm³ at 298K, and the sample B is in a vessel of volume 16.5 dm³ at 298 K. Calculate the number of moles in sample B.
Answer:
n_A = 1.5 mol n_B = ?
V_A = 37.6 dm³ V_B = 16.5 dm³
(T = 298 K constant)

Question 49.
Sulphur hexafluoride is a colourless, odourless gas; calculate the pressure exerted by 1.82 moles of the gas in a steel vessel of volume 5.43 dm³ at 69.5°C, assuming ¡deal gas behaviour.
Answer:
n = 1.82 mole
V = 5.43 dm³
T = 69.5 + 273 = 342.5
P = ?
PV = nRT
P = \(\frac {nRT}{V}\)

P = 94.25 atm.

Question 50.
Argon is an inert gas used in light bulbs to retard the vapourlzation of the tungsten filament. A certain light bulb containing argon at 1.2 atm and 18°C is heated to 85°C at constant volume. Calculate its final pressure in atm.
Answer:
P₁ = 1.2 atm
T₁ = 18°C + 273 = 291 K
T₂ = 85°C + 273 = 358 K
P₂ = ?
\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)

Question 51.
A small bubble rises from the bottom of a lake, where the temperature and pressure are 6°C and 4 atm. to the water surface, where the temperature is 25°C and pressure is 1 atm. Calculate the final volume in (mL) of the bubble, if its initial volume is 1.5 mL.
Answer:
T₁ = 6°C + 273 = 279 K
P₁ = 4 atm V₁ = 1.5m
T₂ = 25°C + 273 = 298 K
P₂ = 1 atm V ₁ = ?
\(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\)

V₂ = 6.41 mol.

Question 52.
Hydrochloric acid is treated with a metal to produce hydrogen gas. Suppose a student carries out this reaction and collects a volume of 154.4 x 10^-3 dm³ of a gas at a pressure of 742 mm of Hg at a temperature of 298 K. What mass of hydrogen gas (in mg) did the student collect?
Answer:
V = 154.4 x 10^-3dm³
P = 742 mm of Hg
T = 298 K m = ?
n = \(\frac{PV}{RT}\) =
= 0.006 mol
n = \(\frac{PV}{RT}\)
n = \(\frac{Mass}{Molar Mass}\)
Mass = n x Molar mass
= 0.006 x 2.016
= 0.0121 g = 12.1 mg.

Question 53.
It takes 192 sec for an unknown gas to diffuse through a porous wall and 84 sec for N2 gas to effuse at the same temperature and pressure. What is the molar mass of the unknown gas?
Answer:

Question 54.
A tank contains a mixture of 52.5 g of oxygen and 65.1 g of CO2 at 300 K the total pressure in the tank ¡s 9.21 atm. Calculate the partial pressure (in atm.) of each gas in the mixture.
Answer:
m_O2 = 52.5 g
P_O2 = ?
m_CO2 = 65.1 g
P_CO2 = ?
T = 300 K P = 9.21 atm
P_O2 = X_O2 x total pressure

P_O2 = X_O2 x Total pressure
= 0.53 x 9.21 atm = 4.88 atm
P_CO2 = X_CO2 x Total pressure
= 0.47 x 9.21 atm = 4.33 atm

Question 55.
A combustible gas Is stored in a metal tank at a pressure of 2.98 atm at 25 °C. The tank can withstand a maximum pressure of 12 atm after which it will explode. The building in which the tank has been stored catches fire. Now predict whether the tank will blow up first or start melting? (Melting point of the metal = 1100 K).
Answer:
T₁ = 298 K;
P₁ = 2.98 atm;
T₂ = 1100K;
P₂ = ?
\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)
P₂ = \(\frac{P_{1}}{T_{1}} \times T_{2}\)
= \(\frac{2.98 \mathrm{atm}}{298 \mathrm{K}} \times 1100 \mathrm{K}\) = 11 atm
At 1100 K, the pressure of the gas inside the tank will become 11 atm. Given that tank can withstand a maximum pressure of 12 atm, the tank will start melting first.

In Text Questions – Evaluate Yourself

Question 1.
Freon-I 2, the compound widely used in the refrigerator system as coolant causes depletion of ozone layer. Now It has been replaced by eco-friendly compounds. Consider 1.5 dm³ sample of gaseous freon at a pressure of 0.3 atm. If the pressure is changed to 1.2 atm. at a constant temperature, what will be the volume of the gas increased or decreased?
Answer:
Volume of freon (V₁) = 1.5 dm³
Pressure (P₁) = 0.3 atm
‘T’ is constant
P₂ = 1.2 atm
V₂ = ?
P₁V₁ = P₂V₂
V₂ = \(\frac{P_{1} V_{1}}{P_{2}}\)

= 0.375 dm³
Volume decreased from 1.5 dm³ to 0.375 dm³

Question 2.
Inside a certain automobile engine, the volume of air in a cylinder is 0.375 dm³, when the pressure is 1.05 atm. When the gas is compressed to a volume of 0.125 dm³ at the same temperature, what is the pressure of the compressed air?
Answer:
V₁ = 0.375dm³
V₂ = 0.125 dm³
P₁ = 1.05 atm
P₂ = ?
T – Constant
P₁V₁ = P₂V₂
P₂ = \(\frac{P_{1} V_{1}}{V_{2}}=\frac{10.5 \times 0.375}{0.125}\)
= 3.15 atm.

Question 3.
A sample of gas has a volume of 3.8 dm³ at an unknown temperature. When the sample is submerged in ice water at 0°C, its volume gets reduced to 2.27 dm³. What is its initial temperature?
Answer:
V₁ = 3.8 dm³
T₂ = 0°C = 273K
T₁ = ? V₂ = 2.27dm³
\(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\)

T₁ = 457 K

Question 4.
An athlete in a kinesiology research study has his lung volume of 7.05 dm³ during a deep inhalation. At this volume the lungs contain 0.312 mole of air. During exhalation the volume of his Jung decreases to 2.35 dm³ How many moles of air does the athlete exhale during exhalation? (assume pressure and temperature remain constant)
Answer:
V₁ = 7.05 dm³
V₂ = 2.35 dm³
n₁ = 0.312 mol
n₁ =?
‘P’ and ‘T’ are constant
\(\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}\)

n₂ = 0.104 mol
Number of moles exhaled = 0.312 – 0.104 = 0.208 moles

Question 5.
A small bubble rises from the bottom of a lake, where the temperature and pressure are 8° C and 6.4 atm. to the water surface, where the temperature ¡s 25°C and pressure is 1 atm. Calculate the final volume in (ml) of the bubble, if its initial volume is 2.1 ml.
Answer:
T₁ = 8°C = 8 + 273 = 281K
P₁ = 6.4atm V₁ = 2.1 mol
T₂ = 25°C = 25 + 273 = 298 K
P₂ = 1 atm
V₂ = ?
\(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\)

V₂ = 14.25 ml

Question 6.
(a) A mixture of He and O₂ were used ¡n the ‘air’ tanks of underwater divers for deep dives. For a particular dive 12 dm³ of O₂ at 298 K, I atm. and 46 dm3 of He, at 298 K, 1 aim. were both pumped into a 5 dm³ tank. Calculate the partial pressure of each gas and the total pressure In the tank at 298 K
Answer:

P = 1 atm
V_total = 5 dm³
P_O2 = X_O2 x P_total

P_He = 3.54 atm

Question 6.
(b) A sample of solid KClO₃ (potassium chlorate) was heated in a test tube to obtain O₂ according to the reaction 2KClO₃ → 2KCl_(s) + 3O₂ The oxygen gas was collected by downward displacement of water at 295 K. The total pressure of the mixture is 772 mm of Hg. The vapour pressure of water is 26.7 mn of Hg at 300K. What is the partial pressure of the oxygen gas?
Answer:
2KCl_3(s) → 2KCl_(s) 3O_3(g)
P_total = 772 mm Hg
P_H2O = 26.7 mm Hg
P_total = P_O2 + P_H2O
P_O2 = P_total – P_H2O
P₁ = 26.7 mm Hg
T₁ = 300 K
T₂ = 295 K
P₂ = ?
\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)

P₂ = 26.26 mm Hg
∴ P_O2 = 772 – 26.26
= 745.74 mm Hg

Question 7.
A flammable hydrocarbon gas of particular volume is found to diffuse through a small hole in 1.5 minutes. Under the same conditions of temperature and pressure an equal volume of bromine vapour takes 4.73 min to diffuse through the same hole. Calculate the molar mass of the unknown gas and suggest what this gas might be, (Given that molar mass of bromine = 159.8 g/ mole)
Answer:
t₁ = 1.5 minutes (gas)_hydrocarbon
t₂ = 4.73 minutes (gas)_Bromi

n (12) + (2n + 2) 1 = 16 (general formula for hydrocarbon C_nH_2n+2)
12n + 2n + 2 = 16
14n = 16 – 2
14n = 14
n = 1
The hydrocarbon is C₁H_{2(1) + 2} = CH₄

Question 8.
Critical temperature of H₂O, NH₃ and CO₂ are 647.4, 405.5 and 304.2 K, respectively. When we start cooling from a temperature of 700 K which will liquefy first and which will liquefy finally?
Answer:
Critical temperature of a gas is defined as the temperature above which it cannot be liquefied even at high pressures.
∴ When cooling starts from 700 K, H₂O vill liquefied first, then followed by ammonia and finally carbon dioxide will liquefied.

In-Text Example Problems

Question 9.
In the below figure, let us find the missing parameters [volume in (b) and pressure in (c)]
P₁ = 1 atm P₂ = 2 atm P₃ = ? atm
V₁ I dm3  V₂ =? dm3 V₃ = 0.25 dm³
T₁ = 298 K T₂ = 298 K T₃ = 298 K
Solution:
According to Boyle’s law, at constant temperature for a given mass of gas at constant temperature,
P₁V₁ = P₂V₂ = P₃V₃
I atm x 1 dm³ = 2 atm x V₂ = P₃x 0.25 dm³
∴ 2 atm x V₂ = 1 atm x 1 dm³

V₂ = 0.5 dm³
and P₃ x 0.25 dm³ = 1 atm x 1 dm³

P₃ = 4atm

Question 10.
In the below figure, let us find the missing parameters [volume in (b) and temperature in (c)]
P₁ = 1 atm P₂ = 1 atm P₃ = 1 atm
V₁ = 0.3dm³ V₂ = ?dm³ V₃ = 0.15dm³
T₁ = 200K T₂ = 300 K T₃ = ? K
Answer:

According to Charles law,


T₃ = 100k

Question 11.
Calculate the pressure exerted by 2 moles of sulphur hexafluoride in a steel vessel of volume 6 dm³ at 70°C assuming It is an ideal gas.
Answer:
We will use the ideal gas equation for this calculation as below:
P = \(\frac {nRT}{V}\)

= 9.39 atm.

Question 12.
A mixture of gases contains 4.76 mole of Ne, 0.74 mole of Ar and 2.5 mole of Xe. Calculate the partial pressure of gases, if the total pressure is 2 atm. at a fixed temperature. Solve this problem using Dalton’s law.
Answer:
P_Ne = X_Ne P_Total

P_Ne = X_Ne P_Total = 0.595 x 2 = 1.19 atm.
P_Ne = X_Ne P_Total = 0.093 x 2 = 0. 186 atm.
P_Ne = X_Ne P_Total = 0.312 x 2 = 0.624 atm.

Question 13.
An unknown gas diffuses at a rate of 0.5 time that of nitrogen at the same temperature and pressure. Calculate the molar mass of the unknown gas.
Answer:

Question 14.
If a scuba diver takes a breath at the surface filling his lungs with 5.82 dm3 of air what volume will the air in his lungs occupy when he drives to a depth, where the pressure ¡s 1.92 atm. (assume temperature is constant and the pressure at the surface is exactly 1 atm.)
Solution :
Temperature = Constant
Pressure at the surface = 1 atm – P₁
Pressure at the depth = 1.92 atm – P₂
Vdlume of air breathing at the surface of the air = 5.82 dm³ – V₁
Volume of air breathing at the depth = V₂ = ?

V₁ = 3.03 dm³
The volume of air scuba diver’s lung occupy = 3.03 dm³

Question 15.
Inside a certain automobile engine, the volume of air in a cylinder is 0.475 dm³, when the pressure is 1.05 atm. When the gas is compressed, the pressure increased to 5.65 atm. at the same temperature. What is the volume of compressed air?
Solution:
Volume of air in the cylinder 0.475 dm³ – V₁
Pressure of air P₁ = 1.05 atm
Increased pressure P₂ = 5.65 atm
Volume of air compressed V₂ = ?
P₁V₁ = P₂V₂

V₂ = 0.08827 dm³
Compressed volume of air = 0.08 827 dm³

Samacheer Kalvi 11th Chemistry Gaseous State Additional Questions Solved

I. Choose the correct answer.

Question 1.
For one mole of a gas, the ideal gas equation is ………..
(a) PV = \(\frac {1}{2}\) RT
(b) PV = RT
(c) PV = \(\frac {3}{2}\)RT
(d) PV = \(\frac {5}{2}\)RT
Answer:
(b) PV= RT

Question 2.
The average kinetic energy of the gas molecule is …………
(a) inversely proportional to its absolute temperature
(b) directly proportional to its absolute temperature
(c) equal to the square of its absolute temperature
(d) All of the above
Answer:
(b) directly proportional to Its absolute temperature

Question 3.
Which of the following is the correct mathematical relation for Charles’ law at constant pressure?
(a) V ∝ T
(b) V ∝ t
(c) V ∝ – \(\frac {1}{T}\)
(d) all of above
Answer:
(a) V ∝ T

Question 4.
At constant temperature, the pressure of the gas is reduced to one-third, the volume
(a) reduce to one-third
(b) increases by three times
(c) remaining the same
(d) cannot be predicted
Answer:
(b) increases by three times

Question 5.
With rise in temperature, the surface tension of a liquid …………
(a) decreases
(b) increases
(c) remaining the same
(d) none of the above
Answer:
(a) decreases

Question 6.
Viscosity of a liquid is a measure of ……………
(a) repulsive forces between the liquid molecules
(b) frictional resistance
(c) intermolecular forces between the molecules
(d) none of the above
Answer:
(b) frictional resistance

Question 7.
The cleansing action of soaps and detergents is due to …………..
(a) internal friction
(b) high hydrogen bonding
(c) viscosity
(d) surface tensions
Answer:
(d) surface tensions

Question 8.
In Vander Waals equation of state for a non-ideal gas the net force of attraction among the molecules is given by ………..
(a) \(\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}\)
(b) P + \(\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}\)
(c) P – \(\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}\)
(d) – \(\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}\)
Answer:
(a) \(\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}\)

Question 9.
The compressibility factor, z for an ideal gas is ………….
(a) zero
(b) less than one
(c) greater than one
(d) equal to one
Answer:
(d) equal to one

Question 10.
Which of the following gases will have the lowest rate of diffusion?
(a) H₂
(b) N₂
(c) F₂
(d) O₂
Answer:
(c) F₂

Question 11.
Which of the following is a mono atomic gas in nature?
(a) Oxygen
(b) Hydrogen
(c) Helium
(d) Ozone
Answer:
(c) Helium

Question 12.
Which of the following is a diatomic gas in nature?
(a) Oxygen
(b) Ozone
(c) Helium
(d) Radon
Answer:
(a) Oxygen

Question 13.
Which one of the following is not a monoatomic gas?
(a) Neon
(b) Xenon
(c) Argon
(d) Oxygen
Answer:
(d) Oxygen

Question 14.
Among the following groups which contains monoatomic gases?
(a) Group 17
(b) Group 18
(c) Group 1
(d) Group 15
Answer:
(b) Group 18

Question 15.
Which of the following is a tri atomic gas at room temperature?
(a) Oxygen
(b) Helium
(c) Ozone
(d) Nitrogen
Answer:
(c) Ozone

Question 16.
Which of the following gas is essential for our survival?
(a) N₂
(b) H₂
(c) O₂
(d) He
Answer:
(c) O₂

Question 17.
Among the following, which is deadly poison?
(a) CO₂
(b) HCN
(c) HCl
(d) NH₃
Answer:
(b) HCN

Question 18.
Which of the following is not chemically inert?
(a) Helium
(b) Oxygen
(c) Argon
(d) Krypton
Answer:
(b) Oxygen

Question 19.
Match the List-I and List-II using the correct code given below the list.

Answer:

Question 20.
Pressure of a gas is equal to ………..
(a) \(\frac {F}{a}\)
(b) F x a
(c) \(\frac {a}{F}\)
(d) F – a
Answer:
(a) \(\frac {F}{a}\)

Question 21.
The SI unit of pressure is ………..
(a) Nm^-2  Kg^-1
(b) Pascal
(c) bar
(d) atmosphere
Answer:
(b) Pascal

Question 22.
Statement-I: The pressure cooker takes more time for cooking at high altitude.
Statement-II: Air is subjected to Earth’s gravitational force. The pressure of air gradually decreases from the surface of the Earth to higher altitude.
(a) Statement-I and II are correct and Statement-II is the correct explanation of Statement-I
(b) Statement-I and II are correct but Statement-II is not the correct explanation of Statement-I
(e) Statement-I is wrong but Statement-II is correct
(ð) Statement-I is correct but Statement-II is wrong
Answer:
(a) Statement-I and II are correct and Statement-II is the correct explanation of Statement-I

Question 23.
The instrument used for measuring the atmospheric pressure is …………….
(a) lactometer
(b) barometer
(c) electrometer
(d) ammeter
Answer:
(b) barometer

Question 24.
The standard atmospheric pressure at sea level at 0°C is equal to ……………..
(a) 1 mm Hg
(b) 76 mm Hg
(c) 760 mm Hg
(d) 680 mm Hg
Answer:
(c) 760 mm Hg

Question 25.
Mathematical expression of Boyle’s law is ………….
(a) P₁V₁ = P₂V₂
(b) \(\frac {P}{V}\) = Constant
(c) \(\frac {V}{T}\) = Constant
(J) \(\frac {P}{T}\) = Constant
Answer:
(a) P₁V₁ = P₂V₂

Question 26.
Statement-I: If the volume of a fixed mass of a gas is reduced to half at constant temperature the gas pressure doubles.
Statement-II: If the volume is halved, the density of the gas is doubled.
(a) Statement-I and II are correct and Statement-II is the correct explanation of Statement-I
(b) Statement-I and II are correct but Statement-II is not the correct explanation of Statement-I
(c) Statement-I is correct but Statement-II is wrong
(d) Statement-I is wrong but Statement-II is correct
Answer:
(a) Statement-I and liare correct and Statement-II is the correct explanation of Statement-I

Question 27.
Which one of the following represents the Charles’ law?
(a) PV = Constant
(b) \(\frac {V}{T}\) = Constant
(c) VT Constant
(d) \(\frac {T}{V}\) = R
Answer:
(b) \(\frac {V}{T}\) = Constant

Question 28.
Which one of the following is absolute zero?
(a) 293 K
(b) 273 K
(c) – 273.15°C
(d) 0°C
Answer:
(c) – 273.15°C

Question 29.
\(\frac {P}{T}\) = Constant is known as …………..
(a) Boyle’s law
(b) Charles’ law
(c) Gay Lussac’s law
(d) Dalton’s law
Answer:
(c) Gay Lussac’s law

Question 30.
The ideal gas equation is …………..
(a) PV = RT for 1 mole
(b) P₁ V₁ = P₂V₂
(c) \(\frac {P}{T}\) = R
(d) P = P₂ + P₂ + P₂
Answer:
(a) PV = RT for 1 mole

Question 31.
The value of Universal gas constant in a ideal gas equation is equal to ………….
(a) 8.3 14 KJ
(b) 0.082057 dm³ atm mol^-1 K^-1
(c) 1 Pascal
(d) 8.314 x 10^-2Pascal
Answer:
(b) 0.082057 dm³ atm mol^-1 K^-1

Question 32.
Mathematical expression of Graham’ s law is ……………….

Answer:

Question 33.
Which law is used in the isotopic separation of deuterium and protium?
(a) Boyle’s law
(b) Charles’ law
(c) Graham’ s law
(d) Gay Lussac’s law
Answer:
(c) Graham’ s law

Question 34.
The value of compression factor Z is equal to ………….
(a) \(\frac {nRT}{PV}\)
(b) \(\frac {PV}{RT}\)
(c) PV x nRT
(d) \(\frac {PV}{nRT}\)
Answer:
(d) \(\frac {PV}{nRT}\)

Question 35.
The value of critical volume is equal in terms of Vander Waals constant is ……….
(a) 3b
(b) \(\frac{8a}{27 Rb}\)
(c) \(\frac{a}{27 b^{2}}\)
(d) \(\frac{2a}{Rb}\)
Answer:
(a) 3b

Question 36.
The value of critical temperature of carbon dioxide is …………
(a) 273 K
(b) 303.98 K
(c) 373 K
(d) – 80°C
Answer:
(b) 303.98 K

Question 37.
Match the List-I and List-II using the correct code given below the list.


Answer:

Question 38.
The value of critical pressure of CO₂ is ……………….
(a) 173 atm
(b) 73 atm
(c) 1 atm
(d) 22.4 atm
Answer:
(b) 73 atm

Question 39.
The temperature below which a gas obey Joule Thomson effect is called …………..
(a) critical temperature
(b) standard temperature
(c) inversion temperature
(d) normal temperature
Answer:
(c) Inversion temperature

Question 40.
The substance used in adiabatic process of liquefaction is ……………
(a) liquid helium
(b) gadolinium sulphate
(c) iron sulphate
(d) liquid ammonia
Answer:
(b) Gadolinium sulphate

Question 41.
The temperature produced in adiabatic process of liquefaction is …………..
(a) zero kelvin
(b) -273 K
(c) 10^-4 K
(d) 10⁴ K
Answer:
(c) 10^-4 K

Question 42.
The molecules of a gas A travel four times faster than the molecules of gas B at same temperature. The ratio of molecular weight M_A/ M_B is ………….
(a) 1/16
(b) 4
(c) 1/4
(d) 16
Answer:
(a) 1/16

Question 43.
The compressibility factor for an ideal gas is …………….
(a) 1.5
(b) 2
(c) 1
(d) x
Answer:
(c) 1

Question 44.
Which of the following pair will diffuse at the same rate?
(a) CO₂ and N₂O
(b) CO₂ and NO
(c) CO₂ and CO
(d) N₂O and NO
Answer:
(a) CO₂ and N₂O

Question 45.
The value of Vander Waals constant “a” is maximum for ……………….
(a) helium
(b) nitrogen
(c) methane
(d) ammonia
Answer:
(d) Ammonia

Question 46.
A person living in Shimla observed that cooking food with using pressure cooker takes more time. The reason for this observation is that at high altitude …………..
(a) pressure increases
(b) temperature decreases
(c) pressure decreases
(a) temperature decreases
Answer:
(c) pressure decreases

Question 47.
Statement-I : At constant temperature PV vs V plot for real gases is not a straight line.
Statement-II : At high pressure, all gases have Z >1, but at intermediate pressure most gases have Z <1.
(a) Statement-I and II are correct and Statement-II is the correct explanation of Statement-I
(b) Statement-I and II are correct but Statement-II is not the correct explanation of Statement-I
(c) Statement-I is correct but Statement-Il is wrong
(d) Statement-I is wrong but Statement-Il is correct .
Answer:
(a) Statement-I and liare correct and Statement-II is the correct explanation of Statement-I

Question 48.
Statement-I: Gases do not liquefy above their critical temperature, even on applying high press ure.
Statement-II: Above critical temperature, the molecular speed is high and intermolecular
attractions cannot hold the molecules together because they escape because of high speed.
(a) Statement-I and II are correct and Statement-II is the correct explanation of Statement-I
(b) Statement-I and II are correct but Statement-II is not the correct explanation of Statement-I
(c) Statement-I is correct but Statement-II is wrong
(d) Statement-I is wrong but Statement-II is correct
Answer:
(c) Statement-I and II are correct and Statement-IIs the correct explanation of Statement-I

Question 49.
The rate of diffusion of a gas is ………….
(a) directly proportional to its density
(b) directly proportional to its molecular mass
(c) directly proportional to its square root of its molecular mass
(d) inversely proportional to its square root of its molecular mass
Answer:
(d) inversely proportional to its square root of its molecular mass

Question 50.
In a closed flask of 5 liters, 1.0 g of H₂ is heated from 300 to 600 K, which statement is not correct’?
(a) pressure of the gas increases
(b) the rate of the collusion increase
(c) the number of moles of gas increases
(d) the energy of gaseous molecules increases
Answer:
(c) the number of moles of gas increases

Question 51.
Match the List-I and List-II using the correct code given below the list.

Answer:

Question 52.
Consider the following statements.
(i) All the gases have higher densities than liquids and solids.
(ii) All gases occupy zero volume at absolute zero.
(iii) At very low pressure all gases exhibit ideal behaviour.
Which of the above statement is/are not correct?
(a) (i) only
(b) (ii) only
(c) (iii) only
(d) (ii) and (iii) only
Answer:
(a) (i) only

Question 53.
Which of the following gas is present maximum in atmospheric air?
(a) O₂
(b) N₂
(c) H₂
(d) radon
Answer:
(b) N₂

Question 54.
Which law is used in the process of enriching the isotope of U²³⁵ from other isotopes?
(a) Boyle’s law
(b) Dalton’s law of partial pressure
(c) Graham’s law of diffusion
(d) Charles’ law
Answer:
(c) Graham’s law of diffusion

Samacheer Kalvi 11th Chemistry Gaseous State 2 – Mark Questions

Question 1.
Identify the elements that are in gaseous state under normal atmospheric conditions.
Answer:

• Hydrogen, nitrogen, oxygen, fluorine and chlorine exist as gaseous diatomic molecules.
• Another form of oxygen namely ozone tr iatomic molecule exist as a gas at room temperature.
• Noble gases namely helium, neon, argon, krypton, xenon and radon are mono atomic gases.

Question 2.
Distinguish between a gas and a vapour.
Answer:

• Gas : A substance that is normally in a gaseous state at ordinary temperature and pressure. .,e.g., Hydrogen.
• Vapou r: The gaseous form of any substance that is a liquid or solid at normal temperature and pressure. e.g., At 298 K and 1 atm, water exist as water vapour.

Question 3.
Define pressure. Give its units.
Answer:

• Pressure is defined as the force exerted by a gas on unit area of the wall. Force F
• Pressure = \(\frac {Force}{Area}\) = \(\frac {F}{a}\)
• The SI unit of pressure is Pascal (Pa)

Question 4.
Define atmospheric pressure. What is its value?
Answer:

• The pressure exerted on a unit area of Earth by the colunm of air above it is called atmospheric pressure.
• The standard atmospheric pressure = 1 atm.
• 1 atm = 760 mm Hg.

Question 5.
Deep sea divers ascend slowly and breath continuously by time they reach the surface. Give reason.
Answer:

• For every 10 m of depth, a diver experiences an additional 1 atm of pressure due to the weight of water surrounding him.
• At 20 m, the diver experiences a total pressure of 3 atm. So the most important rule in diving is never hold breath.
• Divers must ascend slowly and breath continuously allowing the regulator to bring the air pressure in their lungs to 1 atm by the time they reach the surface.

Question 6.
Most aeroplanes cabins are artificially pressurized. Why?
Answer:
The pressure decreases with the increase in altitude because there are fewer molecules per unit volume of air. Above 9200 m (30,000 ft), for example, where most commercial aeroplanes fly, the pressure is so low that one could pass out for lack of oxygen. For this reason most aeroplanes cabins arc artificially pressurized.

Question 7.
What is the reason behind the cause of ear pain while climbing a mountain? How it can be rectified?
Answer:

• When one ascends a mountain in a plain, the external pressure drops while the pressure within the air cavities remains the same. This creates an imbalance.
• The greater internal pressure forces the eardrum to bulge outward causing pain.
• With time and with the help of a yawn or two, the excess air within your ear’s cavities escapes thereby equalizing the internal and external pressure and relieving the pain.

Question 8.
State Charles’ law.
Answer:
Charles’ law:
For a fixed mass of a gas constant pressure, the volume is directly proportional to temperature (K).
Mathematically V – T at constant P and n. (or) \(\frac {V}{T}\) = Constant (or) \(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\) = Constant

Question 9.
What are the applications of Charles’ law?
Answer:

• A hot air inside the balloon rises because of its decreased density and causes the balloon to float  inside the balloon rises because of its decreased density and causes the balloon to float.
• If you take a helium balloon outside on a chilly day, the balloon will crumble. Once you get back into warm area, the balloon will return to its original shape. This is because, in accordance with Charles’ law, a gas like helium takes up more space when it is warm.

Question 10.
State Avogadro’s hypothesis.
Answer:
Equal volumes of all gases under the same conditions of temperature and pressure contain equal number of molecules. Mathematically V ∝ n
\(\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}\) = Constant

Question 11.
Define Dalton’s law of partial pressure.
Answer:
Dalton’s law of partial pressure:
It states that the total pressure of a mixture of gases is the sum of partial pressures of the gases present.
P_total = P₁ + P₂ + P₃

Question 12.
What are the applications of Dalton’s law of partial pressure?
Answer:
1. Physicians report the pressure of the patient’s gases in blood, analyzed by hospital lab, the values are reported as partial pressures.
Gas – Normal range
P_CO2 35 – 45mm of Hg
P_O2  80 – 100 mm of Hg

2. When gas is collected by downward displacement of water, the pressure of dry vapour collected is computed using Dalton’s law

Question 13.
How can you identify a heavy smoker with the help of Dalton’s law?
Answer:
Physicians report the pressure of the patient’s gases in blood, analyzed by hospital lab. The values are reported as partial pressures.

A heavy smoker may be expected to have low O₂ and huge CO₂ partial pressures.

Question 14.
Define Graham’s law of diffusion.
Answer:
Graham’s law of diffusion:
The rate of diffusion or effusion is inversely proportional to the square root of molecular mass of a gas through an orifice.
\(\frac{\mathrm{r}_{\mathrm{A}}}{\mathrm{r}_{\mathrm{B}}}=\sqrt{\frac{\mathrm{M}_{\mathrm{B}}}{\mathrm{M}_{\mathrm{A}}}}\)
r_Ar_B = rate of diffusion of gases A, B
M_A, M_B = Molecular mass of gases A, B

Question 15.
Helium diffuses more than air. Give reason.
Answer:
Take two balloons, one is filled with air and another with helium. After one day, the helium balloon was shrunk, because helium being lighter diffuses out faster than the air

Question 16.
Explain about the applications of Graham’s law of diffusion.
Answer:

• Graham’s law of diffusion is useful to determine the molecular mass of the gas if the rate of diffusion is known.
• Graham’s law forms the basis of the process of enriching the isotopes of U²³⁵ from other isotopes and also useful in isotopic separation of deuterium and protium.

Question 17.
What is compression factor?
Answer:
The deviation of real gases from ideal behaviour is measured in terms of a ratio of PV to nRT.
This is termed as compression factor.
Compression factor = Z = \(\frac {PV}{nRT}\)
For ideal gases Z = 1 at all temperature and pressures.

Question 18.

1. Define critical temperature.
2. What is the critical temperature of CO₂ gas?

Answer:

1. The temperature below which a gas can be liquefied by application of pressure is known as critical temperature.
2. The critical temperature of CO₂ gas is 303.98 K.

Question 19.

1. Define critical pressure.
2. What is the critical pressure of CO₂ gas?

Answer:

1. Critical temperature (P_C) of a gas is defined as the minimum pressure required to liquefy.
2. The critical pressure of CO₂ is 73 atm.

Question 20.
CO₂ gas cannot be liquefied at room temperature. Give reason.
Answer:
Only below the critical temperature, by the application of pressure, a gas can be liquefied. CO₂ has critical temperature as 303.98 K. Room temperature means (30 + 273 K) 3O₃ K. At room temperature, (critical temperature) even by applying large amount of pressure CO₂ cannot be liquefied. Only below the critical temperature, it can be liquefied. At room temperature, CO₂ remains as gas.

Question 21.
What is meant by Joule-Thomson effect?
Answer:
The phenomenon of lowering of temperature when a gas is made to expand adiabatically form a region of high pressure into a region of low pressure is known as Joule-Thomson effect.

Question 22.
Define inversion temperature.
Answer:
The temperature below which a gas obey Joule-Thomson effect is called inversion temperature (T_i).
Ti = \(\frac {2a}{Rb}\)

Question 23.
State and explain Boyle’s law. Represent the law graphically.
Answer:
It states that, the pressure of a fixed mass of a gas is inversely proportional to its volume if temperature is kept constant.
P – \(\frac {1}{V}\)
PV = Constant (n and T are constant)
P₁v₁ = P₂V₂
Graphical Representation:

Question 24.
Give an expression for the van der Waals equation. Give the significance of the constants used in the equation. What are their units?
Answer:
\(\left(P+\frac{n^{2} a}{V^{2}}\right)\) (V- nb) = nRT
Where n is the number of moles present and ‘a’’ b’ are known as van der Waals constants.

Significance of Van der Waals constants:
Van der Waals constant ‘a’:
‘a’ is related to the magnitude of the attractive forces among the molecules of a particular gas. Greater the value of’a’, more will be the attractive forces.
Unit of’a’ = L² mol^-2

Van der Waals constant ‘b’:
‘b’ determines the volume occupied by the gas molecules which depends upon size of molecule.
Unit of ‘b’ = L mol^-1

Question 25.
What are ideal and real gases? Out of CO₂ and NH₃ gases, which is expected to show more deviation from the ideal gas behaviour?
Answer:
Ideal gas:
A gas that follows Boyle’s law, Charles’ law and Avogadro law strictly is called an ideal gas. It is assumed that intermolecular forces are not present between the molecules of an ideal gas.

Real gases:
Gases which deviate from ideal gas behaviour are known as real gases. NH₃ is expected to show more deviation. Since NH₃ is polar in nature and it can be liquefied easily.

Samacheer Kalvi 11th Chemistry Gaseous State 3 – Mark Questions

Question 1.
Explain the graphical representation of Boyle’s law.
Answer:
Boyle’s law states that at a given temperature the volume occupied by a fixed mass of a gas is inversely proportional to its pressure. V ∝ (T and n are fixed) .
If the pressure of the gas increases, volume will decrease and if the pressure of the gas decreases, the volume will increase. So PV = Constant.

Question 2.
What are the consequences of Boyle’s law?
Answer:
1. if the volume of a fixed mass of a gas is reduced to half at constant temperature the gas pressure doubles.
2. Boyle’s law also helps to relate pressure to density.
P₁V₁ = P₃V₃ (Boyle’s law)
\(P_{1} \frac{m}{d_{1}}=P_{2} \frac{m}{d_{2}}\)
Where ‘m’ is the mass, d₁ and d₂ are the densities of gases at pressure P₁ and P₂. The density of the gas is directly proportional to pressure.

Question 3.
Explain Charles’ law with an experimental illustration.
Answer:
Charles’ law states that for a fixed mass of a gas at constant pressure, the volume is directly proportional to temperature (K).
V ∝ T (or) \(\frac {V}{T}\) = Constant

Volume vs Temperature:
If a balloon is moved from an ice water bath to a boiling water bath, the gas molecules inside move faster due to increased temperature and hence the volume increases.

Question 4.
Explain the graphical representation of Charles’ law.
Answer:
1. Variation of volume of the gas sample with temperature at constant pressure.
2. Each line (iso bar) represents the variation of volume with temperature at certain pressure. The pressure increases from P₁ to P₅.
3.  i.e. P₁<P₂ < P₃  <P₄ < P₅ . When these lines are extrapolated to zero volume, they intersect at a temperature of -273.15°C.
4. All gases are becoming liquids, if they are cooled to sufficiently low temperatures.
5. In other words, all gases occupy zero volume at absolute zero. So the volume of a gas can be measured over only a limited temperature range.

Question 5.
Explain graphicl representation of Gay Lussac’s law
Answer:
Gay Lussac’s law
At constant volume, the pressure of a fixed mass of a gas is directly proportional to temperature.
P – T (or) \(\frac {P}{T}\) = Constant
It can be graphically represented as shown here:
Lines in the pressure vs temperature graph are known as isochores (constant volume) of a gas.

Question 6.
Explain the graphical representation of Avogadro’s hypothesis.
Answer:
Avogadros hypothesis states that equal volumes of all gases under the same conditions of temperature and pressure contain equaLnumbr of molecules.
V ∝ n
\(\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}\) = Constant.
Where V₁ and are the volume and number 10- of moles of a gas and V₂ and n₂ are the different set of values of volume and number of moles of the same gas.

A better example to iflustrate Avogadro’s Number of Moles (n) hypothesis is to observe the effect of pumping more gas into a balloon. When more gas molecules (particularly CO₂) are passed, the volume of the balloon increases. The pressure and temperature stay constant as the balloon inflates, so the increase in volume is due to the increase in the quantity of gas inside the balloon.

Question 7.
Derive ideal gas equation.
Answer:
The gaseous state is described completely using the following four variables T, P, V and n. Each gas law relates one variable of a gaseous sample to another while the other two variables are held constant. Therefore, combining all equations into a single equation will enable to account for the change in any or all of the variables.
Boyle’s law: V ∝ \(\frac{1}{P}\)
Charles’ law: V ∝ T
Avogadro’s law: V ∝ n
We can combine these equations into the following general equation that describes the physical
behaviour of all gases.
V ∝ \(\frac{nT}{P}\)
V = \(\frac{nRT}{P}\)
V = , where R = Proportionately constant.
The above equation can be rearranged to give PV = nRT – Ideal gas equation. Where, R is also known as Universal gas constant.

Question 8.
Derive the various values of R, gas constant.
Answer:
1. For standard conditions in which P is 1 atm, volume 22.4 14 dm³ for 1 mole at 273.15 K.

= 0.08205 7 dm³ atm mol^-1 K^-1

2. Where P = 10⁵ Pascal, V = 22.71 x 10^-3 m³ for I mole of a gas at 273.15 K.

= 8.314 pa m³ K^-1 mol^-1
=8.314x 10^-2bar dm³ K^-1mol^-1
R = 8.314 J K^-1 mol^-1.

Question 9.
What is meant by Boyle temperature (or) Boyle point? How is it related with compression point?
Answer:
(1) Over a range of low pressures, the real gases can behave ideally at a particular temperature called as Boyle temperature or Boyle point.
(2) The Boyle point varies with the nattirc of the gas.
(3) Above the Boyle point, the compression point Z > 1 for real gases i.e. real gases show positive deviation.
(4) Below the Boyle point, the real gases first show a decrease for Z, reaches a minimum and then increase with increase in pressure. So, it is clear that at low pressure and at high temperature, the real gases behave as ideal gases.

Question 10.
Explain the different methods used for liquefaction of gases.
Answer:

• Linde’s method: Joule-Thomson effect is used to get liquid air or any other gas.
• Claude’s process: In addition to Joule-Thomson effect, the gas is allowed to perform mechanical work so that more cooling is produced.
• Adiabatic process: This method of cooling is produced by removing the magnetic property of magnetic material e.g. Gadolinium sulphate. By this method, a temperature of 10^-4 K i.e. as low as zero Kelvin can be achieved.

Samacheer Kalvi 11th Chemistry Gaseous State 5 – Mark Question

Question 1.
How CH₄ He and NH₃ are deviating from ideal behaviour? (or) Explain how real gases deviate from ideal behaviour.
Answer:
1. The gases which obeys gas equation PV = nRT are known as ideal gases. The gases which do not obey PV = nRT are known as real gases.

2. The gas laws and the kinetic theory are based on the assumption that molecules in the gas phase occupy negligible volume (assumption 1) and that they do not exert any force on one another either attractive or repulsive (assumption 2). Gases whose behaviour is consistent with these assumptions are said to exhibit ideal behaviour.

3. The following graph shows RT plotted against P for three real gases and an ideal gas at a given temperature.

4. According to ideal gas equation, PV/RT is equal to n. Plot PV/RT versus P for ‘n’ moles oigas at 0°C. For1 mole of an ideal gas PV/ RT is equal to 1 irrespective of the pressure of the gas.

5. For real gases, we observe various deviations from ideal behaviour at high pressure. At very low pressure, all gases exhibit ideal behaviour, ie. PV/RT values all converge to n as P approaches zero.

6. For real gases, this is true only at moderate low pressures. (≤ 5 atm) significant variation occurs as the pressure increases. Attractive forces operate among molecules at relatively short deviation.

7. At atmospheric pressure, the molecules in a gas are far apart and attractive forces arc negligible. At high pressure, the density of the gas increases and the molecules are much closer to one another. Intermolecular forces can be significant enough to affect the motion of the molecules and the gas will not behave ideally.

Question 2.
Derive Van der Waals equation of state.
Answer:
1. Consider the effect of intermolecular forces on the pressure exerted by a gas form the following explanation.

2. The speed of a molecule that is moving toward the wall of a container is reduced by the attractive forces exerted by its neighbours. Hence, the measured gas pressure is Q lower than the pressure the gas would exert, if it behave ideally.


Where ‘a’ is the proportionality constant and depends on the nature of the gas and n and V are the number of moles and volume of the container and respectively an²/ V² is the correction term.

3. The frequency of encounters increases with the square of the number of molecules per unit volume n²/ V². Therefore an²/ V² represents the intermolecular interaction that causes non-ideal behaviour.

4.  Another correction is concerned with the volume ¿ccupied by the gas molecules. ‘V’ represents the volume of the container. As every individual molecule of a real gas occupies certain volume, the effective volume V- nb which is the actually available for the gas, n is the number of moles and b is a constant of gas.

5.  Hence Van der Waals equation of state for real gases are given as \(\left(P+\frac{a n^{2}}{V^{2}}\right)\) (V – nb) = nRT Where a and b are Van der Waals constants.

Question 3.
Explain about Andrew’s experimental isotherms of CO₂ gas.
Answer:
1. Andrew plotted isotherms of carbon dioxide at different temperatures. it is then proved that many real gases behave in a similar manner like CO₂.

2. At a temperature of 303.98 K, CO₂ remains as a gas. Below this temperature, CO₂, turns into liquid CO₂ at 7 3atm. It is called the critical temperature of CO₂.

3. At 303.98 K and 73 atm pressure, CO₃,, becomes a liquid but remains a gas at higher temperature.

4. Below the critical temperature, the behaviour of CO₂ is different. For example, consider an isotherm of CO₂ at 294.5 K, it is a gas until the point B, is reached. At B, a liquid separates along the line BC, both the liquid and gas co-exist. At C, the gas is completely condensed.

5. If the pressure is higher than at C, only the liquid is compressed so, a steep rise in pressure is observed. Thus, there exist a continuity of state.

6.  A gas below the critical temperature can be liquefied by applying pressures.

Activity – 1
The table below contains the values of pressure measured at different temperatures for 1 moie of an ideal gas. Plot the values in a graph and verify the Gay Lussac’s law. [Lines in the pressure vs temperature graph are known as iso chores (constant volume) of a gas]

Solution:
Gay Lussac’s law at constant volume = \(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)

If temperature increases, pressure also increases.
So \(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)

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Students who are preparing for the Science exam can download this Tamilnadu State Board Solutions for Class 10th Science Chapter 17 from here for free of cost. These Tamilnadu State Board Textbook Solutions PDF cover all 10th Science Reproduction in Plants and Animals Book Back Questions and Answers.

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Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 17 Reproduction in Plants and Animals

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Samacheer Kalvi 10th Science Reproduction in Plants and Animals Textual Evaluation Solved

I. Choose the Correct Answer.

Reproduction In Plants And Animals Class 10 Samacheer Question 1.
The plant which propagates with the help of its leaves is ______.
(a) Onion
(b) Neem
(c) Ginger
(d) Bryophyllum.
Answer:
(d) Bryophyllum

10th Science 17th Lesson Question 2.
Asexual reproduction takes place through budding in:
(a) Amoeba
(b) Yeast
(c) Plasmodium
(d) Bacteria
Answer:
(b) Yeast

Samacheer Kalvi Guru 10th Science Question 3.
Syngamy results in the formation of ______.
(a) Zoospores
(b) Conidia
(c) Zygote
(d) Chlamydospores.
Answer:
(c) Zygote

Reproduction In Plants And Animals Class 10 Pdf Question 4.
The essential parts of a flower are:
(a) Calyx and Corolla
(b) Calyx and Androecium
(c) Corolla and Gynoecium
(d) Androecium and Gynoecium
Answer:
(d) Androecium and Gynoecium

Samacheer Kalvi 10th Standard Question 5.
Anemophilous flowers have ______.
(a) Sessile stigma
(b) Small smooth stigma
(c) Coloured flower
(d) Large feathery stigma.
Answer:
(d) Large feathery stigma.

10th Class Biology Reproduction Lesson Question 6.
Male gametes in angiosperms are formed by the division of:
(a) Generative cell
(b) Vegetative cell
(c) Microspore mother cell
(d) Microspore
Answer:
(a) Generative cell

Question 7.
What is true of gametes?
(a) They are diploid
(b) They give rise to gonads
(c) They produce hormones
(d) They are formed from gonads.
Answer:
(d) They are formed from gonads.

Question 8.
A single highly coiled tube where sperms are stored, get concentrated and mature is known as:
(a) Epididymis
(b) Vasa efferentia
(c) Vas deferens
(d) Seminiferous tubules
Answer:
(a) Epididymis

Question 9.
The large elongated cells that provide nutrition to developing sperms are ______.
(a) Primary germ cells
(b) Sertoli cells
(c) Leydig cells
(d) Spermatogonia.
Answer:
(b) Sertoli cells

Question 10.
Estrogen is secreted by:
(a) Anterior pituitary
(b) Primary follicle
(c) Graffian follicle
(d) Corpus luteum
Answer:
(c) Graffian follicle

Question 11.
Which one of the following is an IUCD?
(a) Copper – T
(b) Oral pills
(c) Diaphragm
(d) Tubectomy.
Answer:
(a) Copper – T

II. Fill in the Blanks.

Question 1.
The embryo sac in a typical dicot at the time of fertilization is ______.
Answer:
Female Gametophyte.

Question 2.
After fertilization the ovary develops into ______.
Answer:
Fruit.

Question 3.
Planaria reproduces asexually by ______.
Answer:
Regeneration.

Question 4.
Fertilization is ______ in humans.
Answer:
Internal.

Question 5.
The implantation of the embryo occurs at about ______ day of fertilization.
Answer:
6 to 7.

Question 6.
_______ is the first secretion from the mammary gland after childbirth.
Answer:
Colostrum.

Question 7.
Prolactin is a hormone produced by ______.
Answer:
Anterior Pituitary.

III. Match the following.

Question 1.

Answer:
1. (b) Amoeba
2. (c) Yeast
3. (a) Spirogyra.

Question 2.
Match the following terms with their respective meanings.

Answer:
1. (c) Delivery of baby from a uterus
2. (a) Duration between pregnancy and birth
3. (d) Release of the egg from Graafian follicle
4. (b) Attachment of zygote to the endometrium.

IV. State whether the following statements are True or False. Correct the false statement.

Question 1.
The stalk of the ovule is called pedicle.
Answer:
False.
Correct Statement: Stalk of the ovule is called funiculus.

Question 2.
Seeds are the product of asexual reproduction.
Answer:
False.
Correct Statement: Seeds are the product of Sexual reproduction.

Question 3.
Yeast reproduces asexually by means of multiple fission.
Answer:
False.
Correct Statement: Yeast reproduces asexually by means of budding.

Question 4.
The part of the pistil which serves as a receptive structure for the pollen is called as style.
Answer:
False.
Correct Statement: The part of the pistil which serves as a receptive structure for the pollen is called as stigma.

Question 5.
Insect pollinated flowers are characterized by dry and smooth pollen.
Answer:
False.
Correct Statement: Insect pollinated flowers are characterized by larger and spiny pollen.

Question 6.
Sex organs produce gametes, which are diploid.
Answer:
False.
Correct Statement: Sex organs produce gametes, which are haploid.

Question 7.
LH is secreted by the posterior pituitary.
Answer:
False.
Correct Statement: LH is secreted by the anterior pituitary.

Question 8.
Menstrual cycle ceases during pregnancy.
Answer:
True.

Question 9.
Surgical methods of contraception prevent gamete formation.
Answer:
True.

Question 10.
The increased level of estrogen and progesterone is responsible for menstruation.
Answer:
False.
Correct Statement: The decreased level of estrogen and progesterone is responsible for menstruation.

V. Answer in a word or sentence.

Question 1.
If one pollen grain produces two male gametes, how many pollen grains are needed to fertilize 10 ovules?
Answer:
One sperm fuses with the egg and forms a diploid zygote. So 10 pollen grains are needed to fertilize 10 ovules.

Question 2.
In which part of the flower germination of pollen grains takes place?
Answer:
Germination of pollen grains takes place in the stigma of the female flower.

Question 3.
Name two organisms which reproduce through budding.
Answer:
Yeast, Hydra.

Question 4.
Mention the function of endosperm.
Answer:
Endosperm provides food to the developing embryo.

Question 5.
Name the hormone responsible for the vigorous contractions of the uterine muscles.
Answer;
Oxytocin, from the posterior pituitary, is responsible for the vigorous contractions of the uterine muscles.

Question 6.
What is the enzyme present in acrosome of sperm?
Answer:
Acrosome contain hyaluronidase an enzyme that help the sperm to enter the ovum during fertilization.

Question 7.
When is World Menstrual Hygiene Day observed?
Answer:
Every year May 28 is observed as World Menstrual Hygiene Day.

Question 8.
What is the need for contraception?
Answer:
Contraception is one of the best birth control measures to check population growth.

Question 9.
Name the part of the human female reproductive system where the following occurs.

1. Fertilization
2. Implantation

Answer:

1. Oviduct of the female genital tract.
2. Uterus

VI. Short Answer Questions.

Question 1.
What will happen if you cut Planaria into small fragments?
Answer:
Breaking of fragments of Planaria results into many fragments. Each fragment having one cell will give rise to a new Planaria, by cell division.

Question 2.
Why is vegetative propagation practiced for growing some type of plants?
Answer:
No gametic fusion is required in vegetative reproduction. In this type, new plantlets are formed from vegetative cells, buds or organ of plant. The vegetative part of plant get detached from the parent body and grows into an Independent daughter plant.

Question 3.
How does binary fission differ from multiple fission?
Answer:

Question 4.
Define Triple fusion.
Answer:
The fusion of one male gamete (n) fuses with the secondary nucleus (2n) to produce primary endosperm nucleus (3n) is called Triple fusion.

Question 5.
Write the characteristics of insect-pollinated flowers.
Answer:
Pollination with the help of insects like flies and honey bees are called Entomophily. To attract those insects, these flowers are brightly coloured, have smell and nectar.

Question 6.
Name the secondary sex organs in male.
Answer:
Secondary sex organs in male are seminiferous tubules, epididymis, sperm duct, seminal vesicles, prostrate gland, cowper’s gland and penis.

Question 7.
What is colostrum? How is milk production hormonally regulated?
Answer:
The first fluid which is released from the mammary gland after childbirth is called colostrum. Milk production from alveoli of the mammary gland is stimulated by prolactin secreted from the anterior pituitary. The ejection of milk is stimulated by the posterior pituitary hormone oxytocin.

Question 8.
How can menstrual hygiene be maintained during menstrual days?
Answer:
Menstrual hygiene to be maintained during menstrual days are:

1. Sanitary pads should be changed regularly to avoid infections due to microbes from vagina and sweat from genitals.
2. Use of warm water to clean genitals helps to get rid of menstrual cramps.
3. Wearing loose clothing rather than tight-fitting clothes will ensure the airflow around the genitals and prevent sweating.

Question 9.
How does developing embryo gets its nourishment inside the mother’s body?
Answer:
The placenta allows the exchange of food materials, diffusion of oxygen, excretion of nitrogenous wastes and elimination of carbon-di-oxide. A cord, containing blood vessels that connect the placenta with the foetus is called the umbilical cord.

Question 10.
Identify the parts A, B, C and D.

Answer:
The parts A, B, C and D are:

Question 11.
Write the events involved in the sexual reproduction of a flowering plant.
(a) Discuss the first event and write the types.
(b) Mention the advantages and the disadvantages of that event.
Answer:
(a) The first event is pollination. Pollination is the process of transfer of pollen grains from anther to stigma of a flower. The two types of pollination are self-pollination and cross pollination.

(b) Advantages of self-pollination:

1. Self-pollination is possible in certain bisexual flowers.
2. Flowers do not depend on agents for pollination.
3. There is no wastage of pollen grains.

Disadvantages of self-pollination:

1. The seeds are less in numbers.
2. The endosperm is minute. Therefore, the seeds produce weak plants.
3. New varieties of plants cannot be produced.

Advantages of cross pollination:

1. The seeds produced as a result of cross pollination, develop and germinate properly and grow into better plants, i.e., cross pollination leads to the production of new varieties.
2. More viable seeds are produced.

Disadvantages of cross-pollination:

1. Pollination may fail due to distance barrier.
2. More wastage of pollen grains.
3. It may introduce some unwanted characters.
4. Flowers depend on the external agencies for pollination.

Question 12.
Why are the human testes located outside the abdominal cavity? Name the pouch in which they are present?
Answer:
The testicles, produce sperm and testosterone. The testicles are located outside the body because the sperms develop best at a temperature, several degrees lower than normal body temperature. The pouch, is scrotum, a sac – like structure, in which the testes are present.

Question 13.
Luteal phase of the menstrual cycle is also called the secretory phase. Give reason.
Answer:
In leutal phase LH and FSH decreases, corpus luteum produces progesterone and its level increases followed by a decline progesterone also stimulates the glands in the uterus to secrete substances that maintain the endometrium and keep it from breaking down. For this reason, this phase of menstrual cycle is called secretory phase.

Question 14.
Why are family planning methods not adopted by all the people of our country?
Answer:

• Illiteracy
• Emphasis is in rural areas and not in villages.
• Door to door campaign to encourage families could not be done because of overpopulation.
• Poor economic status and poverty of most of the people in India.
• Age-Old cultural norms continue to cause, poor family planning practices, all across the country.

VII. Long Answer Questions.

Question 1.
With a neat labelled diagram describe the parts of a typical angiosperms ovule.
Answer:
The main part of the ovule is the nucellus which is enclosed by two integuments, leaving an opening called micropyle. The ovule is attached to the ovary wall by a stalk, called funiculus. The basal part is Chalaza.

The embryo sac contains seven cells and the eighth nuclei located within the nucellus. Three cells at the micropylar end form the egg apparatus and the three cells at the chalazalnte9uments end are the antipodal cells. The remaining two nuclei are called polar nuclei found in the centre. In the egg apparatus, one is the egg cell (female gamete) and the remaining two cells are the synergids.

Question 2.
What are the phases of the menstrual cycle? Indicate the changes in the ovary and uterus.
Answer:
The four phases of the menstrual cycle are:

• Menstrual or Destructive phase
• Follicular or Proliferative phase
• Ovulatory phase
• Luteal or secretory phase.

Events of the menstrual cycle and changes in ovary and changes in the uterus.

VIII. Higher Order Thinking Skills (HOTS) Questions

Question 1.
In angiosperms the pollen germinates to produce pollen tube that carries two gametes. What is the purpose of carrying two gametes when single gamete can fertilize the egg?
Answer:
In angiosperms, one sperm cell fuses with the egg cell to form the zygote, while the other fuses with the two polar nuclei that form the endosperm which nourishes the developing embryo.

Question 2.
Why the menstrual cycle does not take place before puberty and during pregnancy?
Answer:
The reproducing period of a women’s life starts and becomes functional and an increase in sex hormone production starts only in puberty. So the menstrual cycle does not take place before puberty. The release of a mature egg, maintains the lining of the uterus, during pregnancy. During pregnancy, the placenta produces progesterone. This maintains the lining of the uterus during pregnancy and it means that menstruation does not happen.

Question 3.
Read the following passage and answer the questions that follow.
Rahini and her parents were watching a television programme. An advertisement flashed on the screen which was promoting use of sanitary napkins. Rahini’s parents suddenly changed the channel, but she objected to her parents and explained the need and importance of Such advertisement.

(a) What is first menstruation called? When does it occur?
Answer:
The first menstruation is called Menarche. In human females, the menstrual cycle starts at the age of 11 – 13 years which marks the onset of puberty.

(b) List out the napkin hygiene measures taken during menstruation?
Answer:

1. The sanitary pad and tampons should be wrapped properly and discarded because they can spread infections.
2. Sanitary pad or tampon should not be flushed down the toilet.
3. Napkin incinerators are to be used properly for disposal of used napkins.

(c) Do you think that Rahini’s objection towards her parents was correct? If so, Why?
Answer:
Yes. Awareness to be created in maintaining menstrual hygeine and importance of menstrual hygeine for good health.

Samacheer Kalvi 10th Science Reproduction in Plants and Animals Additional Questions Solved

I. Fill in the blanks.

Question 1.
The reproduction process is to preserve individual species and is called ______.
Answer:
Self – perpetuation.

Question 2.
The _______ grains are produced in the anther within the pollen sac.
Answer:
Pollen.

Question 3.
The _____ contains the future plant or embryo, which develops into a seedling.
Answer:
Seed.

Question 4.
Each testis is covered with a layer of fibrous tissue called ______.
Answer:
Tunica albuginea.

Question 5.
The first fluid which is released from the mammary gland after childbirth is called ______.
Answer:
Colostrum.

Question 6.
The organs of the reproductive system are divided into ______ and ____ (accessory) sex organs.
Answer:
Primary, secondary.

Question 7.
The plasma membrane of an ovum is surrounded by inner thin _____ and an outer thick ______.
Answer:
Zona pellucida, Corona Radiata.

Question 8.
______ is the practice of healthy living and personal cleanliness.
Answer:
Hygiene.

Question 9.
Testosterone initiates the process of ______.
Answer:
Spermatogenesis.

Question 10.
The cortex of ovary is composed of a connective tissue called ______.
Answer:
Stroma.

Question 11.
In plants, the fusion of one sperm with the egg is called ______.
Answer:
Syngamy.

Question 12.
The _______ provides energy for the movement of the tail in sperm, causing sperm motility, which is essential for fertilization.
Answer:
Mitochondria.

Question 13.
Self – pollination is also known as ______.
Answer:
Autogamy

Question 14.
During spore formation, a structure called sporangium develops from the ______.
Answer:
Fungal hyphae.

II. Choose the correct pair from the following.

Question 1.
(a) Chalaza and pollination
(b) Calyx and spores
(c) Anemophily and hydrophily
(d) Stamens and Planaria.
Answer:
(c) Anemophily and hydrophily

Question 2.
(a) Oogenesis and spermatogenesis
(b) Fragmentation and Fertilization
(c) Primary follicle and fission
(d) Micropyle and Regeneration.
Answer:
(a) Oogenesis and spermatogenesis

Question 3.
(a) Scrotum and sporangium
(b) Pollination and population
(c) Spermatogenesis and seminiferous tubules
(d) Diaphragm and Blastocyst.
Answer:
(c) Spermatogenesis and seminiferous tubules

Question 4.
(a) Autogamy and Budding
(b) Funiculus and chalaza
(c) Bulbils and spirogyra
(d) Polar nuclei and pollination.
Answer:
(b) Funiculus and chalaza

Question 5.
(a) Tuberous root and granulosa cells
(b) Corolla and Graafian follicle
(c) Synergids and Regeneration
(d) The vegetative and generative cell.
Answer:
(d) The vegetative and generative cell.

III. Match the following.

Question 1.

Answer:

1. (f) Sexual reproduction
2. (g) Pollination by wind
3. (e) Seminiferous tubules
4. (c) Blood vessels
5. (a) Pollination by insects
6. (d) Expulsion of the young one
7. (b) Pollination by animals.

IV. Choose the correct answer.

Question 1.
The male and the female gametes contain this material on the chromosomes, which transmit the character traits to the next generation.
(a) genes
(b) chromoplast
(c) septum
(d) spores
Answer:
(a) genes

Question 2.
Raphe and hilium in seed represent:
(a) Nucellus
(b) Funiculus
(c) Integument
(d) Micropyle
Answer:
(b) Funiculus

Question 3.
The corona radiata is formed of ______.
(a) Vitelline membrane
(b) Oxytocin
(c) Zygote
(d) Follicle.
Answer:
(d) Follicle.

Question 4.
Pollination is followed by:
(a) Seed formation
(b) Fragmentation
(c) Fertilization
(d) Budding
Answer:
(c) Fertilization

Question 5.
The membrane forming the surface layer of the ovum is called ______.
(a) Entomophily
(b) Exine
(c) Vitelline membrane
(d) Intine.
Answer:
(c) Vitelline membrane

V. Write True or False for the following statements. Write the correct statement for the incorrect statement.

Question 1.
Sexual reproduction involves the fusion of two diploid gametes to form a haploid individual (Zygote).
Answer:
False.
Correct Statement: Sexual reproduction involves the fusion of two haploid gametes, to form a diploid individual (Zygote).

Question 2.
Accessory sex organ in man includes the Gonads. (Testes in males and Ovaries in female).
Answer:
False.
Correct Statement: Primary sex organ in man include the Gonads. (Testes in males and Ovaries in female).

Question 3.
The connective tissue of Cortex called stroma is lined by the germinal epithelium cells called Granulosa cells.
Answer:
True.

Question 4.
The mature ovum or egg in the female is elliptical in shape and has full of yolk.
Answer:
False.
Correct Statement: The mature ovum or egg is spherical in shape and is always free of yolk.

Question 5.
The menstrual cycle ceases around 48-50 years of age and this stage is termed as menopause.
Answer:
True.

VI. Answer the following briefly.

Question 1.
What is the role of acrosome in human sperm?
Answer:
The sperm head has a like structure called acrosome. It contain hyaluronidase an enzyme that helps the sperm to enter the ovum during fertilization.

Question 2.
What are the significance of fertilization and the post-fertilization changes?
Answer:
Significance of fertilization:

• It stimulates the ovary to develop into a fruit.
• It helps in the development of new characters from two different individuals.

Post – fertilization changes:

• The ovule develops into a seed.
• The integuments of the ovule develop into the seed coat.
• The ovary enlarges and develops into a fruit.

Question 3.
What is Asexual reproduction? Explain the spore formation in Rfiizoptis with a diagram.
Answer:
Production df an offspring by a single parent without the formation and fusion of gametes is called Asexual reproduction. Asexual reproduction occurs by spore formation. In Rhizopus, during spore formation, a structure called sporangium develops from the fungal hypha. The nucleus divides several times within the sporangium and each nucleus with a small amount of cytoplasm develops into a spore. The spores are liberated and they develop into new hypha after reaching the ground or substratum.

Question 4.
What could be the reasons for adopting contraceptive methods?
Answer:
Contraception is one of the best birth control measures. A number of techniques or method have been developed to prevent pregnancies in women which leads to welfare of the family group and society.

Question 5.
Explain the following:
(a) Toilet Hygiene
(b) Napkin Hygiene
(c) Menstrual Hygiene
Answer:
(a) Toilet Hygiene:

• To reduce the bad odour and infection, the floors of the toilet should be kept clean and dry.
• Toilet flush handles, doorknobs, light switches and walls should be cleaned with disinfectants to kill harmful germs and bacteria.
• Hands should be washed with soap, before and after toilet use.

(b) Napkin Hygiene:

• To prevent infections, the sanitary pad, and tampon should be wrapped and discarded properly.
• Sanitary pad and tampon should not be flushed down the toilet.
• Napkin incinerators are to be used properly for disposal of used napkins.

(c) Menstrual Hygiene:

• Sanitary pads should be changed regularly, to avoid infections due to microbes from vagina and sweat from genitals.
• Use of warm water to clean genitals helps to get rid of menstrual cramps.
• Wear loose clothing, to have airflow around the genitals and prevent sweating.

Question 6.
Explain the structure of the following:
(a) Testes
(b) Ovary
Answer:
(a) Testes: Testes are the reproductive glands of the male, that are Oval shaped, which lie outside the abdominal cavity of a man, in a sac-like structure called Scrotum. Each testis is covered with a layer of fibrous tissue called tunica albuginea. Septa divides the testes into pyramidal lobules, in which lie seminiferous tubules, cells of Sertoli and the Leydig cells. The process of spermatogenesis takes place in the seminiferous tubules. The Sertoli cells are the supporting cells and provide nutrients to the developing sperms. The Leydig cells lie between the seminiferous tubules and secrete testosterone. It initiates the process of spermatogenesis.

(b) Ovary: Ovaries are located on either side of the lower abdomen, composed of two almond-shaped bodies, each lying near the lateral end of the fallopian tube. Each ovary has an outer cortex and an inner medulla. The cortex is composed of a network of connective tissue called stroma and is lined by the germinal epithelium. The epithelial cells called the granulosa cells to surround each ovum in the ovary together forming the primary follicle. As the egg grows larger, the follicle also enlarges and gets filled with the fluid and is called the Graafian follicle.

Question 7.
What do you know about the National Health Programme?
Answer:
To improve the reproductive health of the people, National Health Programmes such as,
(a) Family Welfare Programme:

• Maternal and Child Health Care (MCH).
• Immunization of mothers, infants and children.
• Nutritional supplement to pregnant women and children.
• Motivate couples to accept contraceptive methods and to have small family norms, which improve economic status, living status and quality of life.

(b) Reproductive and Child Health Care (RCH)

• Pregnancy and childbirth.
• Postnatal care of the mother and child.
• Importance of breastfeeding.
• Prevention of reproductive tract infections and sexually transmitted diseases.

Question 8.
What is gastrulation?
Answer:
The transformation of blastula into gastrula and the formation of primary germ layers (Ectoderm, Mesoderm and Endoderm) by rearrangement of the cells is called gastrulation.

Question 9.
What is implantation?
Answer:
The process of attachment of the blastocyst to the uterine wall (endometrium) is called implantation. The blastocyst (fertilized egg) reaches the uterus and gets implanted in the uterus. The fertilized egg becomes implanted in about 6 to 7 days after fertilization.

VII. Draw a labelled diagram of the following.

Question 1.
Cross Section of:
(a) Human testes
(b) Human ovary
Answer:
(a) Human testes:

(b) Human ovary:

VIII. Write the expansion for the following abbreviations.

Question 1.

1. WHO
2. RCH
3. MCH
4. STD
5. IUD
6. UTI
7. LH
8. FSH.

Answer:

1. WHO – World Health Organization
2. RCH – Reproductive and Child Health Care
3. MCH – Maternal and Child Health Care
4. STD – Sexually Transmitted Diseases
5. IUD – Intra-Uterine Device
6. UTI – Urinary Tract Infection
7. LH – Luteinizing Hormone
8. FSH – Follicle Stimulating Hormones.

IX. Answer the following in detail. Draw diagrams wherever necessary.

Question 1.
With the help of a neat labelled diagram, explain, how does fertilization take place in flowering plants.
Answer:
The mature pollen grain contains two cells, the vegetative and the generative cell. The vegetative cell contains a large nucleus. The generative cell divides mitotically and forms two male gametes.

Pollen grains reach the stigma and begin to germinate. Pollen grain forms a small tube-like structure called pollen tube, which emerges through the germ pore. The contents of the pollen grain move into the tube. The pollen tube grows through the tissues of the stigma and style and finally reaches the ovule through the micropyle.

The vegetative cell degenerates and the generative cell divides to form two sperms or male gametes. Tip of pollen tube bursts and the two sperms enter the embryo sac. One sperm fuses with the egg and this process is called Syngamy and forms a diploid zygote. The other sperm fuses with the secondary nucleus, which is called the Triple fusion, to form the primary endosperm nucleus, which is triploid in nature.

Since two types of fusion syngamy and triple fusion take place in an embryo sac, the process is termed as double fertilization. After triple fusion, primary endosperm nucleus develops into an endosperm, which provides food to the developing embryo. Later the synergids and antipodal cells degenerate.

Question 2.
With examples and with the help of a neat labelled diagram, explain the different types of vegetative reproduction of plants.
Answer:
Vegetative reproduction takes place through,
(i) Leaves:

In Bryophyllum, small plants grow at the leaf notches.

(ii) Stem:
In plants like, strawberry aerial weak stems, touch the ground and give adventitious roots and buds. The offspring becomes an independent plant, when it is detached from the parent plant.

(iii) Root: Tuberous roots are used to develop new plants, eg. Asparagus, Sweet potato.

(iv) Bulbils: In some plants, the flower bud, modifies into the globose bulb, which is called bulbils. When these bulbils fall on the ground they grow into new plants, eg. Agave.

(v) Fragmentation: The breaking of the filament into many fragments is called fragmentation. Each fragment having one cell will give rise to a new filament by cell division, eg. Spirogyra.

(vi) Fission:

The parent cell divides into two daughter cells and each cell develops into a new adult organism eg. Amoeba.

(vii) Budding:

Formation of a daughter individual from a small projection, the bud arising on the parent body is called budding, eg. Yeast.

(viii) Regeneration: The ability of the lost body parts of an individual organism to give rise to a whole new organism is called regeneration. Regeneration takes place by the specialized mass of cells, eg. Hydra and Planaria.

Question 3.
Explain the two types of pollination and write the advantages and disadvantages of the types of pollination.
Answer:
The two types of pollination are
(a) Self – pollination: The transfer of pollen grains from the anther to the stigma of the same flower or another flower borne on the same plant is called self – pollination.
eg. Hibiscus. Self – pollination is also called Autogamy.
Advantages of self – pollination:

• Self – pollination is possible in certain bisexual flowers.
• Flowers do not depend on agents for pollination.
• There is no wastage of pollen grains.

Disadvantages of self – pollination:

• The seeds are less in numbers.
• The endosperm is minute, so the seeds produce weak plants.
• New varieties of plants cannot be produced.

(b) Cross-pollination: The transfer of pollen grains from the anthers of a flower to the stigma of a flower on another plant of the same species is called Cross-pollination, eg. Apples, Grapes, Plum.
Advantages of Cross-pollination:

• The seeds produced as a result of cross-pollination, develop and germinate properly and grow into better plants.
• Cross-pollination leads to the production of new varieties.
• More viable seeds are produced.

Disadvantages of Cross-pollination:

• Pollination may fail due to the distance barrier.
• More wastage of pollen grains.
• It may introduce some unwanted characters.
• Flowers depend on the external agencies for pollination.

Question 4.
Explain the structure of Human Sperm and Ovum with a neat labelled diagram.
Answer:
1. Structure of Human sperm:
The spermatozoa consist of a head, middle piece and tail. The sperm head is elongated and formed by the condensation of the nucleus. The anterior portion has a cap-like structure called acrosome, which contains an enzyme hyaluronidase, that helps the sperm to enter the ovum during fertilization. A short neck connects the head and middle piece, which comprises the centrioles. The middle piece contains the mitochondria which provides energy for the movement of tail. It brings about sperm motility, which is essential for fertilization.

2. Structure of Ovum:
The mature ovum or egg is spherical in shape, free of yolk and contains abundant cytoplasm and the nucleus. The ovum is surrounded by three membranes. The plasma membrane is surrounded by inner thin zona pellucida and an outer thick corona radiata, which is formed of follicle cells. The membrane forming the surface layer of the ovum is called the vitelline membrane. The fluid-filled space between zona pellucida and the surface of the egg is called perivitelline space.

Question 5.
Explain the Agents of cross-pollination.
Answer:
Cross – pollination takes place through the agency of animals, insects, wind and water.
(a) Pollination by the wind: The pollination with the help of wind is called Anemophily. The Anemophilous flowers produce an enormous amount of pollen grains, which are small, smooth, dry and light in weight. This kind of pollen can blow off at a distance of more than 1,000 km. The stigmas are large, protruding and hairy to trap the pollen grains, eg. Grasses and some Cacti.

(b) Pollination by insects: Pollination with the help of insects like honey bees, flies are called entomophily. These flowers are brightly coloured, have the smell and have nectar, to attract insects. The pollen grains are larger in size, and the exine is pitted and spiny, so it can easily adhere firmly on the sticky stigma. About 80% of the pollination by insects is carried by honey bees.

(c) Pollination by water: The pollination with the help of water is called hydrophily. This takes place in aquatic plants. Pollen grains are produced in large numbers. Pollen grains float on the surface of water till they land on the stigma of female flowers, eg. Hydrilla, Vallisneria.

(d) Pollination by Animals: When pollination takes place with the help of animals, it is called Zoophily. The flowers have bright colour, size and scent to attract animals, eg. Sunbird pollinates flowers of Canna, Gladioli etc., Squirrels pollinate flowers of silk cotton tree.

Question 6.
Explain the parts of a flower, with a neat labelled diagram.
Answer:
The flower is the reproductive organ of a flowering plant. A flower is a modified shoot. A flower consists of four whorls borne on the thalamus. The whorls are from outside. The parts of a typical flower are as follows:

(a) Calyx: Calyx consists of sepals, forms the outermost whorl. They are non-essential or accessory whorls. It protects the flower bud.

(b) Corolla: Corolla consists of petals, which are modified leaves. They are often brightly coloured and in different shapes to attract pollinators. They are non-essential or accessory whorls because they do not directly take part in the reproduction.

(c) Androecium:

Androecium is the male reproductive part of a flower and is called an essential whorl, as they take part directly in reproduction. The androecium is composed of stamens. Each stamen consists of a stalk called the filament and a small bag like structure called anther at the tip. The pollen grains are produced in the anther within the pollen sac.

Pollen grains are spherical and have a two-layered wall. The hard outer layer is exine, which has apertures, called Germpore. The inner thin layer is intine, which is continuous and made up of cellulose and pectin. Mature pollen grain contain two cells, the vegetative and generative cell. Vegetative cell contains a large nucleus. The generative cell divides, to form two male
gametes.

(d) Gynoecium:
Gynoecium is the female reproductive part of a flower. It is an essential whorl because they take part directly in reproduction. Gynoecium is made up of carpels. It has (a) Ovary (b) Style (c) Stigma. Ovary contains the ovules. The main part of the ovule is the nucellus and is enclosed by two integuments, Jk. leaving an opening called as micropyle. The ovule is attached to the ovary wall, by a stalk called Funiculus.

Question 7.
Explain the steps involved from fertilization to foetal development in human.
Answer:
Fertilization in human is internal and occurs in the Oviduct of the female genital tract. It takes place in the ampulla of the fallopian tube. An oocyte is alive for about 24 hours after it is released from follicle. The sperm enters into the ovum, fuses and forms a zygote, which is a fertilized ovum. This process is called fertilization.

(a) Cleavage and formation of Blastula: The first cleavage takes place about 30 hours after fertilization. Cleavage is a series of rapid mitotic divisions of the zygote, to form many-celled blastula (Blastocyst), which comprises an outer layer of smaller cells and inner mass of larger cells.

(b) Implantation: The blastocyst (fertilized egg) reaches the uterus and gets implanted. The process of attachment of the blastocyst to the uterine wall is called Implantation. The fertilized egg, implanted in about 6 to 7 days after fertilization.

(c) Gastrulation: After implantation, the transformation of blastula into Gastrula and the formation of primary germ layers (Ectoderm, Mesoderm and Endoderm) by rearrangement of the cells takes place, which is called Gastrulation.

(d) Organogenesis: The various organs of the foetus are established from the different germ layers, during Organogenesis.

(e) Formation of Placenta: Placenta is a disc-shaped structure, attached to the Uterine wall between the developing embryo and maternal tissues. It allows the exchange of food materials, diffusion of oxygen, excretion of nitrogenous wastes and elimination of carbon-di-oxide. The cord which connects the placenta with the foetus is called the umbilical cord.

(f) Pregnancy (Gestation): Embryo attains its development in the uterus. Gestation period of human is 280 days. During pregnancy the uterus expands up to 500 times of its normal size.

(g) Parturition (Child Birth): Oxytocin from the posterior pituitary stimulates the uterine contractions and provides force to expel the baby from the uterus, causing birth. Parturition is the expulsion of young one from the mother’s Uterus at the end of gestation.

(h) Lactation: The process of milk production after childbirth from the mammary glands of the mother is called lactation. Milk production from the mammary gland is stimulated by prolactin, a hormone secreted by anterior pituitary. The ejection of milk is stimulated by posterior pituitary hormone oxytocin.

Question 8.
What is population explosion? Explain the different ways of family planning.
Answer:
The sudden and rapid rise in the size of the human population is called the Population Explosion. Contraception: Contraception is one of the best birth control measures. The devices used for contraception are called contraceptive devices. The common contraceptive methods used to prevent pregnancy are as follows:

(a) Barrier methods: This method prevents the meeting of an ovum by the sperms. The entry of sperm is prevented into the female reproductive tract by a barrier.

• Condom: Condom is made of thin rubber or latex sheath. Condom prevents deposition of sperms in the vagina. A condom protects against Sexually Transmitted Diseases (STD) like Syphilis and AIDS.
• Diaphragm (Cervical cap): Vaginal diaphragm fitting into the vagina or a cervical cap fitting over the cervix. This prevents the entry of sperms into the uterus.

(b) Hormonal Methods: Hormonal preparations are in the form of pills or tablets (contraceptive pills). These hormones stop the release of an egg from the ovary.

(c) Intra – Uterine Devices (IUDs): The intrauterine device (IUD) are contraceptive devices, inserted into the uterus. Lippe’s Loop and Copper-T, made of copper and plastic are two synthetic devices, commonly used in India. This can remain for a period of 3 years. This reduces the sperm fertilizing capacity and prevents implantation.

(d) Surgical methods: Surgical contraception or sterilization techniques are terminal methods to prevent any pregnancy. This procedure in males is called as vasectomy (ligation of vas deferens) and in females, it is called tubectomy (ligation of fallopian tube). These are methods of permanent birth control.

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You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.3

11th Maths Exercise 1.3 Answers Question 1.
Suppose that 120 students are studying in 4 sections of eleventh standard in a school. Let A denote the set of students and B denote the set of the sections. Define a relation from A to B as “x related toy if the student x belongs to the section y”. Is this relation a function? What can you say about the inverse relation? Explain your answer.
Solution:
(i) A = {set of students in 11^th standard}
B = {set of sections in 11sup>th standard}
R : A ➝ B ⇒ x related to y
⇒ Every students in eleventh Standard must in one section of the eleventh standard.
⇒ It is a function.
Inverse relation cannot be a function since every section of eleventh standard cannot be related to one student in eleventh standard.

Exercise 1.3 Class 11 Maths State Board Question 2.
Write the values of f at – 4, 1, -2, 7, 0 if

Solution:
f(-4) = -(-4) + 4 = 8
f(1) = 1 – 1² = 0
f(-2) = (-2)² – (-2) = 4 + 2 = 6
f(7) = 0
f(0) = 0

11th Maths Exercise 1.3 Question 3.
Write the values of f at -3, 5, 2, -1, 0 if

Solution:
f(-3) = (-3)² – 3 – 5 = 9 – 8 = 1
f(5) = (5)² + 3(5) – 2 = 25 + 15 – 2 = 38
f(2) = 4 – 3 = 1
f(-1) = (-1)² + (-1) – 5 = 1 – 6 = -5
f(0) = 0 – 3 = -3

11th Maths Exercise 1.3 Solutions Question 4.
State whether the following relations are functions or not. If it is a function check for one-to-oneness and ontoness. If it is not a function, state why?
(i) If A = {a, b, c] and/= {(a, c), (b, c), (c, b)};(f: A ➝ A).
(ii) If X = {x, y, z} and/= {(x, y), (x, z), (z, x)}; (f: X ➝ X).
Solution:
(i) f : A ➝ A

It is a function but it is not 1 – 1 and not onto function.

(ii) f : X ➝ X

x ∈ X (Domain) has two images in the co-domain x. It is not a function.

11th Maths Exercise 1.3 Answers Samacheer Question 5.
Let A = {1, 2, 3, 4} and B = {a, b, c, d}. Give a function from A ➝ B for each of the following:
(i) neither one-to-one nor onto.
(ii) not one-to-one but onto.
(iii) one-to-one but not onto.
(iv) one-to-one and onto.
Solution:
A = {1, 2, 3, 4}
B = {a, b, c, d}.

R = {(1, b) (2, b) (3, c) (4, d)} is not 1-1 and not onto

(iii) Not possible

(iv)

11 Maths Exercise 1.3 Question 6.
Find the domain of \(\frac{1}{1-2 \sin x}\)
Solution:

11th Maths Exercise 1.3 In Tamil Question 7.
Find the largest possible domain of the real valued function f(x) = \(\frac{\sqrt{4-x^{2}}}{\sqrt{x^{2}-9}}\)
Solution:

∴ No largest possible domain
The domain is null set

Samacheer Kalvi 11 Maths Solutions Question 8.
Find the range of the function \(\frac{1}{2 \cos x-1}\)
Solution:
The range of cos x is – 1 to 1

Exercise 1.3 Class 11 Maths Question 9.
Show that the relation xy = -2 is a function for a suitable domain. Find the domain and the range of the function.
Solution:
xy = – 2 ⇒ y = -2/x
which is a function
The domain is (-∞, 0) ∪ (0, ∞) and range is R – {0}

Samacheer Kalvi Guru 11th Maths Question 10.
If f, g : R ➝ R are defined by f(x) = |x| + x and g(x) = |x| – x, find gof and fog.
Solution:

Samacheer Kalvi 11th Maths Question 11.
If f, g, h are real valued functions defined on R, then prove that
(f + g)oh = foh + goh. What can you say about fo(g + h) ? Justify your answer.
Solution:
Let f + g = k

= (f + g((h(x))
= f[h(x)] + g [h(x)]
= foh + goh
(i.e.,)(f + g)(o)h = foh + goh
fo(g + h) is also a function

Samacheer Kalvi Guru 11 Maths Question 12.
If f: R ➝ R is defined by f(x) = 3x – 5, prove that f is a bijection and find its inverse.
Solution:
P(x) = 3x – 5
Let y = 3x – 5 ⇒ 3x = y + 5

Samacheer Kalvi 11th Maths Book Back Answers Question 13.
The weight of the muscles of a man is a function of his body weight x and can be expressed as W(x) = 0.35x. Determine the domain of this function.
Solution:
W(x) = 0.35x
Since body weight x is positive and if it increases then W(x) also increase.
Domain is (0, ∞) i.e.,x > 0

Samacheerkalvi.Guru 11th Maths Question 14.
The distance of an object falling is a function of time t and can be expressed as s(t) = -16t². Graph the function and determine if it is one-to-one.
Solution:
s(t) = -16t²
Suppose S(t₁) = S(t₂)

since time cannot be negative, we to take t₁ = t₂
Hence it is one-one.

Samacheer Kalvi.Guru 11th Maths Question 15.
The total cost of airfare on a given route is comprised of the base cost C and the fuel surcharge S in rupee. Both C and S are functions of the mileage m; C(m) = 0.4m + 50 and S(m) = 0.03m. Determine a function for the total cost of a ticket in terms of the mileage and find the airfare for flying 1600 miles.
Solution:
C – base cost,
S = fuel surcharge,
m = mileage
C(m) = 0.4 m + 50
S(m) = 0.03 m
Total cost = C(m) + S(m)
= 0.4 m + 50 + 0.03 m
= 0.43 m + 50
for 1600 miles
T(c) = 0.43 (1600) + 50 = 688 + 50 = ₹ 738

Samacheer Kalvi Class 11 Maths Question 16.
A salesperson whose annual earnings can be represented by the function A(x) = 30, 000 + 0.04x, where x is the rupee value of the merchandise he sells. His son is also in sales and his earnings are represented by the function S(x) = 25, 000 + 0.05x. Find (A + S)(x) and determine the total family income if they each sell Rupees 1,50,00,000 worth of merchandise.
Solution:
A(x) = 30, 000 + 0.04x, where x is merchandise rupee value
S(x) = 25000 + 0.05 x
(A + S) (x) = A(x) + S(x)
= 30000 + 0.04x + 25000 + 0.05 x
= 55000 + 0.09x
(A + S) (x) = 55000+ 0.09x
They each sell x = 1,50,00,000 worth of merchandise
(A + S) x = 55000 + 0.09 (1,50,00,000)
= 55000 + 13,50,000
∴ Total income of family = ₹ 14,05,000

Samacheer Kalvi Class 11 Maths Solutions Question 17.
The function for exchanging American dollars for Singapore Dollar on a given day is f(x) = 1.23x, where x represents the number of American dollars. On the same day the function for exchanging Singapore Dollar to Indian Rupee is g(y) = 50.50y, where y represents the number of Singapore dollars. Write a function which will give the exchange rate of American dollars in terms of Indian rupee.
Solution:
f(x) = 1. 23x where x is number of American dollars.
g(y) = 50.50y where y is number of Singapore dollars.
gof(x) = g(f(x))
= g(1. 23x)
= 50.50 (1.23x)
= 62.115 x

Question 18.
The owner of a small restaurant can prepare a particular meal at a cost of Rupees 100. He estimates that if the menu price of the meal is x rupees, then the number of customers who will order that meal at that price in an evening is given by the function D(x) = 200 – x. Express his day revenue, total cost and profit on this meal as functions of x.
Solution:
cost of one meal = ₹ 100
Total cost = ₹ 100 (200 – x)
Number of customers = 200 – x
Day revenue = ₹ (200 – x) x
Total profit = day revenue – total cost
= (200 – x) x – (100) (200 – x)

Question 19.
The formula for converting from Fahrenheit to Celsius temperatures is \(y=\frac{5 x}{9}-\frac{160}{9}\)
Find the inverse of this function and determine whether the inverse is also a function.
Solution:

Question 20.
A simple cipher takes a number and codes it, using the function f(x) = 3x – 4. Find the inverse of this function, determine whether the inverse is also a function and verify the symmetrical property about the line y = x (by drawing the lines).
Solution:
f(x) = 3x – 4
Let y = 3x – 4

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.3 Additional Questions

Question 1.
Find the domain and range of the function \(f(x)=\frac{1}{\sqrt{x-5}}\)
Solution:
Given that : f(x) \(f(x)=\frac{1}{\sqrt{x-5}}\)
Here, it is clear that / (x) is real when x – 5 > 0 ⇒ x > 5
Hence, the domain = (5, ∞)
Now to find the range put

For x ∈ (5, ∞), y ∈ R⁺.
Hence, the range of f = R⁺.

Question 2.
If \(f(x)=\frac{x-1}{x+1}\), then show that

Solution:

Question 3.
Find the domain of each of the following functions given by:
\(f(x)=\frac{x^{3}-x+3}{x^{2}-1}\)
Solution:

Here, f(x) is not defined if x² – 1 ≠ 0
(x – 1) (x + 1) ≠ 0
x ≠ 1, x ≠ -1
Hence, the domain of f = R – {-1, 1}

Question 4.
Find the range of the following functions given by
f(x) = 1 + 3 cos 2x
Solution:
Given that: f(x) = 1 + 3 cos 2x
We know that -1 ≤ cos 2x ≤ 1
⇒ -3 ≤ 3 cos 2x ≤ 3 ⇒ -3 + 1 ≤ 1 + 3 cos 2x ≤ 3 + 1
⇒ -2 ≤ 1 + 3 cos 2x ≤ 4 ⇒ -2 ≤ f(x) ≤ 4
Hence the range of f = [-2, 4]

Question 5.
Find the domain and range of the function \(f(x)=\frac{x^{2}-9}{x-3}\)
Solution:

Domain off: Clearly f(x) is not defined for x – 3 = 0 i.e. x = 3.
Therefore, Domain (f) = R – {3}
Range off: Let f(x) = y. Then,

It follows from the above relation that y takes all real values except 6 when x takes values in the set R – {3}. Therefore, Range (f) = R {6}.

Question 6.
Find the range of the following functions given by f(x) = \(\frac{1}{2-\sin 3 x}\)
Solution:

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Ex 2.2

9th Maths Exercise 2.2 Samacheer Kalvi Question 1.
Express the following rational numbers into decimal and state the kind of decimal expansion.
(i) \(\frac { 2 }{ 7 }\)
(ii) \(-5 \frac{3}{11}\)
(iii) \(\frac { 22 }{ 3 }\)
(iv) \(\frac { 327 }{ 200 }\)
Solution:
(i) \(\frac { 2 }{ 7 }\)

\(\frac{2}{7}=0 . \overline{285714}\)
Nen-terminating and recurring

(ii) \(-5 \frac{3}{11}\)

\(-5 \frac{3}{11}=-5 . \overline{27}\)
Nen-terminating and recurring

(iii) \(\frac { 22 }{ 3 }\)

\(\frac{22}{3}=7 . \overline{3}\)
Nen-terminating and recurring

(iv) \(\frac { 327 }{ 200 }\)

\(\frac { 327 }{ 200 }\) = 1.635, Terminating.

9th Maths Exercise 2.2 In Tamil Question 2.
Express \(\frac { 1 }{ 13 }\) in decimal form. Find the length of the period of decimals.
Solution:

\(\frac{1}{13}=0 . \overline{076923}\) has the length of the period of decimals = 6.

9th Standard Maths Exercise 2.2 Question 3.
Express the rational number \(\frac { 1 }{ 13 }\) in recurring decimal form by using the recurring decimal expansion of \(\frac { 1 }{ 11 }\) . Hence write \(\frac { 71 }{ 33 }\) in recurring decimal form.
Solution:
The recurring decimal expansion of \(\frac { 1 }{ 11 }\) = 0.09090909…. = \(0.\overline { 09 }\)

9th Maths Exercise 2.2 Question 4.
Express the following decimal expression into rational numbers.
(i) \(0.\overline { 24 }\)
(ii) \(2.\overline { 327 }\)
(iii) -5.132
(iv) \(3.1\overline { 7 }\)
(v) \(17.\overline { 215 }\)
(vi) \(-21.213\overline { 7 }\)
Solution:
(i) \(0.\overline { 24 }\)
Let x = \(0.\overline { 24 }\) = 0.24242424……… ….(1)
(Here period of decimal is 2, multiply equation (1) by 100)
100x = 24.242424 ………. ….(2)
(2) – (1)
100x – x = 24.242424…. – 0.242424….
99x = 24
x = \(\frac { 24 }{ 99 }\)

(ii) \(2.\overline { 327 }\)
Let x = 2.327327327…… …………. (1)
(Here period of decimal is 3, multiply equation (1) by 1000)
1000x = 2327.327… ……………. (2)
(2) – (1)
1000x – x = 2327.327327… – 2.327327….
999x = 2325
x = \(\frac { 2325 }{ 999 }\)

(iii) -5.132
\(x=-5.132=\frac{-5132}{1000}=\frac{-1283}{250}\)

(iv) \(3.1\overline { 7 }\)
Let x = 3.1777 ……. ………… (1)
(Here the repeating decimal digit is 7, which is the second digit after the decimal point, multiply equation (1) by 10)
10x = 31.7777 …….. …………. (2)
(Now period of decimal is 1, multiply equation (2) by 10)
100x = 317.7777…….. …………….. (3)
(3) – (2)
100x – 10x = 317.777…. – 31.777….
90x = 286
\(x=\frac{286}{90}=\frac{143}{45}\)

(v) \(17.\overline { 215 }\)
Let x = 17.215215 ……. ………. (1)
1000x = 17215.215215…… …………. (2)
(2) – (1)
1000x – x = 17215.215215… – 17.215…
999x = 17198
x = \(\frac { 17198 }{ 999 }\)

(vi) \(-21.213\overline { 7 }\)
Let x = -21.2137777… ……….. (1)
10x = -212.137777…… ……….. (2)
100x = -2121.37777…… ………… (3)
1000x = -21213.77777…. ……….. (4)
10000x = 212137.77777….. ………… (5)
(Now period of decimal is 1, multiply equation (4) it by 10)
(5) – (4)
10000x – 1000x = (-212137.7777…) – (-21213.7777…)
9000x = -190924
x = –\(\frac { 190924 }{ 9000 }\)

9th Maths 2.2 Question 5.
Without actual division, find which of the following rational numbers have terminating decimal expansion.
(i) \(\frac { 7 }{ 128 }\)
(ii) \(\frac { 21 }{ 15 }\)
(iii) 4\(\frac { 9 }{ 35 }\)
(iv) \(\frac { 219 }{ 2200 }\)
Solution:
(i) \(\frac { 7 }{ 128 }\)

So \(\frac{7}{128}=\frac{7}{2^{7} 5^{0}}\)
This of the form 4m, n ∈ W
So \(\frac { 7 }{ 128 }\) has a terminating decimal expansion.

(ii) \(\frac { 21 }{ 15 }\)

So \(\frac { 21 }{ 15 }\) has a terminating decimal expansion.

(iii) 4\(\frac { 9 }{ 35 }\) = \(\frac { 149 }{ 35 }\)

\(\frac{49}{35}=\frac{149}{5^{1} 7^{1}}\)
∴ This is not of the form \(\frac{p}{5^{1} 7^{1}}\)
So 4\(\frac { 9 }{ 35 }\) has a non-terminating recurring decimal expansion.

(iv) \(\frac { 219 }{ 2200 }\)

\(\frac{219}{2200}=\frac{219}{2^{3} 5^{2} 11^{1}}\)
∴ This is not of the form \(\frac{p}{2^{m} 5^{n}}\)
So \(\frac { 219 }{ 2200 }\) has a non-terminating recurring decimal expansion.

You can Download Samacheer Kalvi 9th Social Science Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Social Science History Solutions Chapter 5 The Classical World

The Classical World Textual Exercise

I. Choose the correct answer.

The Classical World 9th Class Question 1.
…………….. is the Greek city-state which resisted the Persians to the end.
(a) Acropolis
(b) Sparta
(c) Athens
(d) Rome
Answer:
(c) Athens

Attempt An Account Of Slavery In Rome Question 2.
The other name for Greeks was …………….
(a) Hellenists
(b) Hellenes
(c) Phoenicians
(d) Spartans
Answer:
(b) Hellenes

Samacheer Kalvi 9th Books Social Science Question 3.
The founder of Han dynasty was …………..
(a) Wu Ti
(b) Hung Chao
(c) Liu Pang
(d) Mangu Khan
Answer:
(c) Liu Pang

Question 4.
…………… was the Roman Governor responsible for the crucifixion of Jesus.
(a) Innocent I
(b) Hildebrand
(c) Leo I
(d) Pontius Pilate
Answer:
(d) Pontius Pilate

Question 5.
The Peloponnesian War was fought between ……………… and ……………
(a) Greeks and Persians
(b) Plebeians and Patricians
(c) Spartans and Athenians
(d) Greeks and Romans
Answer:
(c) Spartans and Athenians

II. Find out the correct statement.

Question 1.
(i) First Persian attack on Greece failed.
(ii) The downfall of Roman Empire is attributed to Julius Caesar.
(iii) The Barbarians who invaded Rome were considered to be culturally advanced.
(iv) Buddhism weakened the Roman Empire.
(a) (i) is correct
(b) (ii) is correct
(c) (ii) and (iii) are correct
(d) (iv) is correct
Answer:
(a) (i) is correct

Question 2.
(i) Euclid developed a model for the motion of planets and stars.
(ii) Romans established a republic after overthrowing Etruscans.
(iii) Acropolis became a famous slave market.
(iv) Rome and Carthage united to drive out the Greeks.
(a) (i) is correct
(b) (ii) is correct
(c) (ii) and (iv) are correct
(d) (iv) is correct
Answer:
(c) (ii) and (iv) are correct

Question 3.
(i) Silk road was closed during the Han dynasty.
(ii) Peasant uprisings posed threats to Athenian democracy.
(iii) Virgil’s Aeneid glorified Roman imperialism.
(iv) Spartacus killed Julius Caesar. f
(a) (i) is correct
(b) (ii) is correct
(c) (ii) and (iv) are correct
(d) (iii) is correct
Answer:
(d) (iii) is correct

Question 4.
(i) Roman Emperor Marcus Aurelius was a tyrant.
(ii) Romulus Aurelius was the most admired ruler in Roman History.
(iii) Fabius was a famous Carthaginian General.
(iv) Tacitus is respected more than Livy as a historian.
(a) (i) is correct
(b) (ii) is correct
(c) (ii) and (iii) are correct
(d) (iv) is correct
Answer:
(d) (iv) is correct

Question 5.
(i) Buddhism went to China from Japan
(ii) After crucifixion of Jesus, St Thomas spread the Christian doctrine
(iii) St Sophia Cathedral was the most magnificent building in Europe
(iv) Trajan was one of the worst dictators Rome had.
(a) (i) is correct
(b) (ii) is correct
(c) (iii) is correct
(d) (iv) is correct
Answer:
(c) (iii) is correct

III. Match the following:

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

IV. Fill in the blanks.

1. Greeks defeated the Persians at ………………
2. …………… stood in favour of poor peasants in Roman republic.
3. Buddhism came to China from India during the reign of …………… dynasty.
4. The most magnificent building in Europe was ……………
5. ……….. and …………… were Magistrates in Rome.
Answers:
1. Marathon in 490 B.C.
2. Tiberius Gracchus and Garius do Gracchus
3. Han
4. St. Sophia Cathedral
5. Marius and Sulla

V. Answer all questions given under each heading.

Question 1.
Emergence of Rome as an empire
(a) Who were the Gracchus brothers?
Answer:
Tiberius Gracchus and Garius do Gracchus were Patricians. They voiced their opinion in favour of the poor peasants.

(b) What role did they play?
Answer:
They voiced their opinion in favour of the poor peasants.

(c) What was the outcome of their martyrdom?
Answer:
The Martyrdom of the Gracchus brothers played a decisive role in the transformation of the Roman Republic into the Roman Empire.

(d) Who was the first Roman Emperor?
Answer:
Marus Aureillus was the first Roman Emperor.

Question 2.
Han Dynasty
(a) Who was the founder of Han Empire?
Answer:
Han Dynasty was founded by Liu Pang.

(b) What was the capital of Han Empire?
Answer:
The capital of Han Empire was Chang-an.

(c) Where did they have their new capital?
Answer:
They had their new capital at Xuchang.

(d) Who was the powerful ruler of the Han dynasty?
Answer:
The most popular and powerful ruler was Wu Ti.

VI. Answer the following briefly.

Question 1.
Attempt an account of slavery in Rome.
Answer:

• A major source of revenue to the Roman state was slave trade.
• The island of Delos became a great slave market.
• There were more slave revolts in Rome than in Greece.
• The revolt of spartacus was the most famous.

Question 2.
Highlight the main contribution of Constantine.
Answer:
The main contribution of Constantine was, conversion to Christianity. He himself became a Christian and Christianity became the official religion of the Empire.

Question 3.
What do you know of the Carthaginian leader Hannibal?
Answer:

• Hannibal was a Carthaginian General who defeated the Roman Army and made a great part of Italy a desert in the Punic War.
• In the Second Punic War Hannibal was defeated in the Battle of Zama.
• Pursued by the Roman army, Hannibal ended his life by poisoning himself.

Question 4.
What were the reasons for the prosperity of Han Empire?
Answer:

1. The Han Empire threw open the silk road for trade.
2. A large export trade mainly in silk reached as far as the Roman Empire.
3. In the North artisans and herders of rival “barbarian” dynasties brought in new techniques like the methods harnessing horses, use of saddle and stirrup, techniques of building bridges and mountain roads and seafaring.
4. Such innovations made Han Empire prosperous.

Question 5.
Write about St. Sophia Cathedral.
Answer:

• St. Sophia Cathedral was built in mid-sixth century AD (CE) The most magnificent building in Europe at that time, it was known for its innovative architectural techniques.
• This Cathedral was turned into a mosque by the Ottoman Turks when they captured Constantinople.

VII. Answer the following in detail:

Question 1.
Discuss the rise and growth of Athens, pointing out its glorious legacy.
Answer:
In Athens, the pressure from below resulted in the replacement of both oligarchy’and tyranny by “democracy”. The law-making power in Athens was vested in an assembly open to all freemen. Judges and lower officials were chosen by lots. This arrangement was resented by ‘ the upper classes who considered democracy to be the rule of the mob.

The Persian danger had united the Greeks. When this danger was removed, they started quarrelling again. The history of many Greek city-states was one of continual struggles by the rich landowners against “democracy”. The only exception was Athens, where “democracy’ survived for about 200 years.

Question 2.
Highlight the contributions of Rome to World Civilization.
Answer:
The Byzantine emperors, who ruled from the city of Constantinople for about 1,000 years,
called themselves Romans. But their language was Greek. The splendor of Constantinople with its luxurious royal palaces, its libraries, its scholars familiar with the writings of Greeks 1 and Romans and its fascinating St. Sophia Cathedral are the legacies they have left behind.

However, in terms of the development of science and technology, there was no progress during this I period. The economies of the Empire’s provinces were in the hands of large local landowners. The small peasants always lived on the edge of poverty. The fundamental weakness of
Byzantine Civilization stood exposed when the participants of Fourth Crusade pillaged it and I ruled it.

Student Activities

Question 1.
In an outline map of Europe, the students are to sketch the extent of Western and Eastern Roman Empire.
Answer:

The Classical World Additional Questions

I. Choose the correct answer.

Question 1.
………….. a fortified city of ancient Greeks on a hill in Athens is an illustrative example of their advancement.
(a) Acropolis
(b) Athens
(c) Sparta
(d) None of the above
Answer:
(a) Acropolis

Question 2.
The word …………….. literally means “rule of the people”.
(a) Autocracy
(b) Communism
(c) Socialism
(d) Democracy
Answer:
(d) Democracy

Question 3.
Aristotle was the disciple of ……………
(a) Socrates
(b) Plato
(c) Pericles
(d) Augustus
Answer:
(b) Plato

Question 4.
“Natural History” was completed by ………….
(a) Pliny the Elder
(b) Antoninus Pius
(c) Marcus Aurelius
(d) Tacitus
Answer:
(a) Pliny the Elder

Question 5.
……………. came to China from India during the reign of Han Dynasty.
(a) Jainism
(b) Sikhism
(c) Buddhism
(d) Christianity
Answer:
(c) Buddhism

Question 6.
……………. started spreading the Christian doctrine after the Crucifixion of Jesus.
(a) St. Paul
(b) St. Thomas
(c) St. Antony
(d) St. John
Answer:
(a) St. Paul

II. Find out the correct statement.

Question 1.
(i) Until 8th Century B.C. (BCE) Greece was different from the rest of the world.
(ii) Democracy literally means “rule of the people”.
(iii) The Greek city-states have an elaborate bureaucracy.
(iv) The entire period of Alexander’s reign was not spent on wars.
(a) (i) is correct
(b) (ii) is correct
(c) (ii) and (iii) are correct
(d) (iv) is correct
Answer:
(b) (ii) is correct

Question 2.
(i) Prisoners of war were enslaved in Rome.
(ii) Rome developed into a normal town.
(iii) In the beginning Rome was a society of Business men.
(iv) A major source of revenue to the Roman state was slave trade.
(a) (i) is correct
(b) (ii) is correct
(c) (i) and (iv) are correct
(d) (iii) is correct
Answer:
(c) (i) and (iv) are correct

Question 3.
(i) The revolt of spartacus was the most famous.
(ii) The Island of Delos became a great slave market.
(iii) Catalina’s victory led to mob violence.
(iv) Livy was a Poetist.
(a) (i) is correct
(b) (i) and (ii) are correct
(c) (iii) is correct
(d) (iv) is correct
Answer:
(b) (i) and (ii) are correct

Question 4.
(i) Han Empire once again threw open the silk road for trade.
(ii) Buddhism came to China from Indonesia.
(iii) With Buddhism came the influence of Indian art to Korea.
(iv) The period after Han rule witnessed political stability across the country.
(a) (i) is correct
(b) (i) and (iii) are correct
(c) (ii) is correct
(d) (iii) and (iv) are correct
Answer:
(a) (i) is correct

Question 5.
(i) Jewish had no faith on Jesus.
(ii) They hoped Messiah would not arrive.
(iii) Jesus was against the rich and the hypocrites.
(iv) Christianity did not spread in Europe.
(a) (i) is correct
(b) (ii) is correct
(c) (iii) is correct
(d) (iv) is correct
Answer:
(c) (iii) is correct

III. Match the following:

Answer:
1. (f)
2. (g)
3. (a)
4. (b)
5. (c)
6. (d)
7. (e)

IV. Fill in the blanks.

1. Under ……….. the new rich exploited the smaller landholders.
2. The Persian danger had united the ……………..
3. Cultural development that took place rapidly after Alexander’s death. 323 BC is called ………….. civilization.
4. A major source of revenue to the Roman state was ……………
5. The most distinguished writers of the ……………… Age brought glory to the empire.
6. The most popular and powerful ruler of Han Dynasty was …………….
7. Some of the Buddhist art of the time show the impact of …………… styles.
8. One of the Roman emperors …………….. himself became a Christian.
9. Internal Crisis and invasion of Franks, Goths and Vandals ended the …………….
10. Christianity became a state religion of …………….. and began to spread in Europe.
Answers:
1. Oligarchy
2. Greeks
3. Hellenistic
4. Slave trade
5. Augustus
6. Wu Ti
7. Hellenistic
8. Constantine
9. Roman Empire
10. Byzantium

V. Answer all questions given under each heading.

Question 1.
Athenian Democracy.
(a) What replaced the oligarchy and tyranny in Athens?
Answer:
Democracy.

(b) With whom the law-making power was vested in Athens?
Answer:
The law making power in Athens was vested in an assembly open to all freemen.

(c) What united the Greeks?
Answer:
The Persian danger had united the Greeks.

(d) How many years were Democracy survived in Athens?
Answer:
Democracy survived in Athens for about 200 years.

Question 2.
Hellenistic Civilization.
(a) Who established the kingdom in Macedonia?
Answer:
Alexander the Great.

(b) How did this kingdom succeed?
Answer:
This kingdom succeeded in annexing two historic empires of Egypt and the Middle East.

(c) How did Alexander spend his life?
Answer:
The entire period of Alexander’s reign was spent on wars.

(d) When did the cultural development take place? What is it called?
Answer:
After the death of 323 B.C. and is called Hellenistic civilization.

VI. Answer the following briefly.

Question 1.
What do you know about the classical world?
Answer:

1. Greece, Rome and China represented the classical world which ended with the fall of Western Roman Empire.
2. Until the end of 5th Century AD (CE) Christianity was confined to the Roman Empire.

Question 2.
Why did the first Persian attack on Greece failed?
Answer:
The first Persian attack on Greece failed due to the fact that Persian army suffered from disease and lack of food during its March.

Question 3.
Comment on the ‘Age of Pericles’.
Answer:

1. Athens, despite hostility and disturbance from sparta, became a noble city with magnificent buildings.
2. There were great artists and great thinkers.
3. Historians therefore call this the Age of Pericles.

Question 4.
What do you know about the Socrates?
Answer:

1. Socrates was the greatest of the thinkers of the Pericles Era.
2. As a philosopher he discussed difficult problems with his friends to bring out the truth out of discussions.

Question 5.
Who were the capable rulers in Rome after the death of Augustus?
Answer:
Trajan (98-117), Antoninus Pius (138-161) and Marcus Aurelius (161-180) were the capable enlightened rulers.

Question 6.
Who were Barbarians?
Answer:
A group of people from a very different country (or) culture that is considered to be less culturally advanced and more violent than their own were Barbarians.

Question 7.
What is known as Silk Road (or) Silk Route?
Answer:
The trade route from China to Asia Minor and India, known as the Silk Road or Silk Route, linked China with the West. Goods and ideas between the two great civilizations of Rome and China were exchanged through this route. Silk went westward, and wools, gold, and silver went east. China received Buddhism from India via the Silk Road.

Question 8.
Who spreaded Christianity after Jesus’s crucifixion?
Answer:

1. St. Paul started spreading the Christian doctrine.
2. Paul succeeded in his effort and Christianity gradually spread.

VII. Answer the following in detail.

Question 1.
Explain the “Rise of Christianity”.
Answer:
(i) After a brief period of glory in the days of David and Solomon, the Jewish people had a great fall and experienced extreme hardship.

(ii) While spreading out all over the Roman Empire and elsewhere, they hoped that a Messiah would arrive to restore their pristine glory.

(iii) Initially they laid much hopes on Jesus. Jesus was against the rich and the hypocrites, and condemned certain observances and ceremonials.

(iv) This was not to the liking of the priests, who turned against Jesus and handed him over to the Roman Governor Pontius Pilate. Looked upon as a political rebel by the Roman authorities, Jesus was tried and crucified.

(v) After Jesus’s crucifixion, St Paul started spreading the Christian doctrine. Paul succeeded in his effort and Christianity gradually spread.

(vi) Romans were prepared to tolerate Christianity. But the refusal of the Christians to pay respect to the Emperor’s image was viewed as political treason.

(vii) It led to the persecution of Christians. Their property was confiscated and they were thrown to the lions.

(viii) Yet the Roman Empire did not succeed in suppressing Christianity. One of the Roman emperors Constantine himself became a Christian. Christianity thus became the official religion of the Empire.

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.4

10th Maths Exercise 5.4 Samacheer Kalvi Question 1.
Find the slope of the following straight lines
(i) 5y – 3 = 0
(ii) \(7 x-\frac{3}{17}\) = 0
Solution:

Exercise 5.4 Class 10 Samacheer Kalvi Question 2.
Find the slope of the line which is
(i) parallel to y = 0.7x -11
(ii) perpendicular to the line x = -11
Solution:
(i) y = 0.7x – 11
line parallel to y = 0.7x – 11 is y = 0.7x + C
If the lines are parallel, slopes are equal
∴ The slope of the required line is 0.7.
(ii) m = tan θ = tan 90°= ∞ undefined.

Ex 5.4 Class 10 Samacheer Question 3.
Check whether the given lines are parallel or perpendicular

Solution:
(i) \(\frac{x}{3}+\frac{y}{4}+\frac{1}{7}\) = 0

10th Maths Exercise 5.4 Question 4.
If the straight lines 12y = -(p + 3)x + 12, 12x – 7y = 16 are perpendicular then find ‘p’.
Solution:

10th Maths Example 5.4 Question 5.
Find the equation of a straight line passing through the point P (-5, 2) and parallel to the line joining the points Q(3, -2) and R(-5, 4).
Solution:

Class 10 Maths Exercise 5.4 Solutions Question 6.
Find the equation of a line passing through ; (6, -2) and perpendicular to the line joining the points (6, 7) and (2, -3).
Solution:
Slope of line joining (6, 7) and (2,-3) is

5 Coordinate Geometry Question 7.
A(-3, 0) B(10, – 2) and C(12, 3) are the vertices of ∆ABC . Find the equation of the altitude through A and B.
Solution:
A(-3, 0), B(10, -2), C(12, 3)
Since AD ⊥ BC



(1), (2) are the required equations of the altitudes through A and B.

5.4 Geometry Question 8.
Find the equation of the perpendicular bisector of the line joining the points A(-4, 2) and B(6, -4).
Solution:
Mid Point AB is

Ex 5.4 Class 10 Question 9.
Find the equation of a straight line through the intersection of lines 7x + 3y = 10, 5x – 4y = 1
and parallel to the line 13x + 5y + 12 = O
Solution:

l₁ passes through the intersecting point.

10th Samacheer Kalvi Maths Coordinate Geometry Question 10.
Find the equation of a straight line through the intersection of lines 5x – 6y = 2, 3x + 2y = 10 and perpendicular to the line 4x – 7y + 13 = 0
Solution:

12th Maths Exercise 5.4 Samacheer Kalvi Question 11.
Find the equation of a straight line joining the point of intersection of 3x + y + 2 = 0 and x – 2y – 4 = 0 to the point of intersection of 7x – 3y = -12 and 2y = x + 3
Solution:

Exercise 5.4 Class 10 Question 12.
Find the equation of a straight line through the point of intersection of the lines 8x + 3+ = 18, 4x + 5+ = 9 and bisecting the line segment joining the points (5, -4) and (-7, 6).
Solution:
The intersecting point of the lines

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Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 13 Structural Organisation of Animals

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Samacheer Kalvi 10th Science Structural Organisation of Animals Textual Evaluation Solved

I. Choose the correct answer.

Structural Organisation Of Animals Class 10 Question 1.
In leech, locomotion is performed by _____.
(a) Anterior sucker
(b) Posterior sucker
(c) Setae
(d) None of the above.
Answer:
(d) None of the above.

10th Science Structural Organisation Of Animals Question 2.
The segments of leech are known as:
(a) Metameres (somites)
(b) Proglottids
(c) Strobila
(d) All the above
Answer:
(a) Metameres (somites)

You can Download Samacheer Kalvi 10th Science Book Back Answers Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Samacheer Kalvi Guru 10th Science Question 3.
Pharyngeal ganglion in leech is a part of _____.
(a) Excretory system
(b) Nervous system
(c) Reproductive system
(d) Respiratory system.
Answer:
(b) Nervous system

Samacheer Kalvi 10th Science Question 4.
The brain of leech lies above the:
(a) Mouth
(b) Buccal Cavity
(c) Pharynx
(d) Crop
Answer:
(c) Pharynx

10th Science Samacheer Kalvi Question 5.
The body of leech has _____.
(a) 23 segments
(b) 33 segments
(c) 38 segments
(d) 30 segments.
Answer:
(b) 33 segments

Question 6.
Mammals are ______ animals.
(a) Cold – blooded
(b) Warm – blooded
(c) Poikilothermic
(d) All the above.
Answer:
(b) Warm – blooded

Question 7.
The animals which give birth to young ones are:
(a) Oviparous
(b) Viviparous
(c) Ovoviviparous
(d) All the above
Answer:
(b) Viviparous

II. Fill in the blanks.

Question 1.
The posterior sucker is formed by the fusion of the _______ segments.
Answer:
Last seven.

Question 2.
The existence of two sets of teeth in the life of an animal is called ______ dentition.
Answer:
Diphyodont.

Question 3.
The anterior end of leech has a lobe-like structure called _____.
Answer:
Sucker.

Question 4.
The blood-sucking habit of a leech is known as _____.
Answer:
Sanguivorous.

Question 5.
______ separate nitrogenous waste from the blood in the rabbit.
Answer:
Nephrons.

Question 6.
_____ spinal nerves are present in the rabbit.
Answer:
37 pairs of.

III. Identify whether the statements are true or false. Correct the false statement.

Question 1.
An anticoagulant present in the saliva of the leech is called heparin.
Answer:
False.
Correct Statement: The anticoagulant present in the saliva of the leech is called Hirudin.

Question 2.
The vas deferens serves to transport the ovum.
Answer:
False.
Correct Statement: The vas deferens serves to transport the sperm.

Question 3.
The rabbit has a third eyelid called tympanic membrane which is movable.
Answer:
False.
Correct Statement: The rabbit has a third eyelid called Nictitating membrane, which is movable.

Question 4.
Diastema is a gap between premolar and molar teeth in rabbit.
Answer:
True.

Question 5.
The cerebral hemispheres of the rabbit are connected by a band of nerve tissue called corpora quadrigemina.
Answer:
False.
Correct Statement: The cerebral hemisphere of the rabbit are connected by a band of nerve tissue called Corpus callosum.

IV. Match the Columns I, II and III correctly.

Question 1.

Answer:

V. Answer in a sentence.

Question 1.
Give the common name of the Hirudinaria granulosa.
Answer:
Indian Cattle Leech.

Question 2.
How does leech respire?
Answer:
The leech respire through skin (diffusion).

Question 3.
Write the dental formula of the rabbit.
Answer:
Dental formula is \(\left(\mathrm{I} \frac{2}{1}, \mathrm{C} \frac{0}{0}, \mathrm{PM} \frac{3}{2}, \mathrm{M} \frac{3}{3}\right)\) in Rabbit, which is written as \(\frac{2033}{1023}\)

Question 4.
How many pairs of testes are present in leech?
Answer:
In leech, there are eleven pairs of testes. One pair in each segment from 12 to 22 segments.

Question 5.
How is diastema formed in rabbit?
Answer:
The diastema is formed in Rabbit, as a gap between the incisors and premolars,

Question 6.
What organs are attached to the two bronchi?
Answer:
Lung, Bronchioles and Alveoli are the Organs, attached to Bronchi.

Question 7.
Which organ acts as suction pump in leech?
Answer:
The muscular pharynx found below the buccal cavity acts as suction pump in leech. It is surrounded by large masses, unicellular glands called salivary glands.

Question 8.
What does CNS stand for?
Answer:
CNS stands for Central Nervous System.

Question 9.
Why are the teeth of a rabbit called heterodont?
Answer:
The teeth of Rabbit are of different types. So it is called heterodont.

Question 10.
How does leech suck blood from the host?
Answer:
In leech, blood is sucked by the muscular pharynx. The salivary secretion is poured in the wound. The saliva contains an active substance called hirudin which prevents the coagulation of the blood.

VI. Short Answer Questions.

Question 1.
Why are the rings of cartilages found in the trachea of a rabbit?
Answer:
Tracheal walls are supported by rings of cartilage, which help in the free passage of air.

Question 2.
List out the parasitic adaptations in leech.
Answer:

1. The suckers are the primary organ of the blood sucking.
2. The blood is sucked by muscular pharynx.
3. Leeches attaches itself to the body of host by Anterior and Posterior ends of the body.
4. The three jaws inside the mouth, causes a painless triradiate or Y shaped incision in the skin of the host.
5. A protein called hirudin is produced in the salivary gland of leech to prevent blood coagulation. Thus, a continuous supply of the blood is maintained.
6. Parapodia and setae are completely absent.
7. Leeches also inject an anaesthetic substances that prevents the host from feeling their bite.
8. In the crop, blood is stored which gives nourishment to the leech for several months. Due to this reason there is no eloborate secretion of the digestive juices and enzymes.

VII. Long Answer Questions.

Question 1.
How is the circulatory system designed in leech to compensate the heart structure?
Answer:
There are no true blood vessels in leeches. The blood vessels are replaced by canals called haemocoelic canals. These canals are filled with haemocoelic fluid. There are four longitudinal canals. One is dorsal lying above the alimentary canal, another is ventral lying below the alimentary canal. The remaining two are lateral lying on either side of the alimentary canal. These four canals are connected together at the posterior end. There is no heart, but the lateral channels serve as a heart by being contractile. They have values inside. The dorsal and ventral channels are non-contractile having no muscular walls.

Question 2.
How does locomotion take place in leech?
Answer:
Locomotion in leech takes place by:

• Looping or Crawling movement: This type of movement is brought about, by the contraction and relaxation of muscles. The two suckers, serve for attachment, during movement on a substratum.
• Swimming movement: Leeches swim very actively and perform undulating movements in the water.

Question 3.
Explain the male reproductive system of the rabbit with a labelled diagram.
Answer:
A pair of Testes, which are ovoid in shape, are enclosed by Scrotal Sac, in the abdominal cavity. Each testis consists of numerous fine tubules called Seminiferous tubules. This network of tubules leads into a coiled tubule called Epididymis, which lead into the sperm duct called vas deferens. The vas deferens join in the urethra, just below the Urinary Bladder. The Urethra runs backwards and passes into the penis.

The Prostate gland, Cowper’s gland and Perineal gland are the three accessory glands, whose secretions are involved in reproduction.

VIII. Higher Order Thinking Skills (HOTS) Questions

Question 1.
Arjun is studying in tenth standard. He was down with fever and went to meet the doctor. As he went to the clinic he saw a patient undergoing treatment for severe leech bite. Being curious, Arjun asked the doctor why leech bite was not felt as soon as it attaches to the skin ? What would have been the reply given by the doctor?
Answer:
The leech makes a wound with the jaws by making a rasping movement. The blood is sucked by the muscular pharynx. The salivary secretion is poured in the wound. They inject an anaesthetic substances that prevents the host from feeling their bite. The saliva contains an active substances called hirudin which prevents the coagulation of the blood.

Question 2.
Shylesh has some pet animals at his home. He has few rabbits too, one day while feeding them he observed something different with the teeth. He asked his grandfather, why is it so? What would have been the explanation of his grandfather?
Answer:
The explanation of the grandfather would have been as follows: Teeth are hard bone – like structures, used to cut, tear and grind the food. There are incisors, canines, premolars and molars teeth are seen. Canines are absent. Something different from the teeth is the gap between incisors and premolar, which is called Diastema. It helps in mastication and chewing the food.

IX. Value-Based Questions.

Question 1.
Leeches do not have an secretion of digestive juices and enzymes -Why?
Answer:
The blood sucked by the leech is stored up in the crop. The blood gets haemolysed in the crop. Then the blood is passed drop by drop into the stomach where it is digested slowly by the peptolytic enzyme. The digested blood is absorbed slowly by the intestine.

Question 2.
How is the digestive system of rabbit suited for a herbivorous mode of feeding?
Answer:
The caecum is a thin – walled sac present at the Junction of the small intestine and large intestine. It contains bacteria, that helps in digestion of cellulose. So the digestive system of Rabbit is suited for a herbivorous mode of feeding.

Samacheer Kalvi 10th Science Structural Organisation of Animals Additional Questions Solved

I. Match the following.

Question 1.

Answer:
1. (c) Metamerism
2. (d) Move-in groups
3. (a) Hirudin
4. (e) Septum
5. (b) Vagina.

II. Choose the correct pair.

Question 1.
The digestion of cellulose in rabbit takes place in:
(a) vemiform appendix
(b) colon
(c) caecum
(d) ileum
Answer:
(c) caecum

Question 2.
(a) Ureter and Dermis
(b) Clitella and Mouth
(c) Vagina and Egg case
(d) Sucker and Small intestine.
Answer:
(c) Vagina and Egg case

Question 3.
Total number of incisors teeth in rabbit is:
(a) 8
(b) 6
(c) 10
(d) 4
Answer:
(b) 6

Question 4.
(a) Mouth and Metamerism
(b) Ganglion and Colon
(c) Intestine and Integument
(d) Expiration and Inspiration.
Answer:
(d) Expiration and Inspiration

Question 5.
(a) Sweat glands and Sebaceous glands
(b) Triradiate and Thoracic
(c) Trachea and Tricuspid
(d) Seminiferous and Sansuivorous.
Answer:
(a) Sweat glands and Sebaceous glands

III. Fill in the Blanks.

Question 1.
In leech, the ______ tissue lies beneath longitudinal muscles and fills the entire colon around the gut.
Answer:
Botryoidal.

Question 2.
_____ teeth are absent in Rabbit.
Answer:
Canine.

Question 3.
Digestion in leech takes place in the stomach by the action of ______ enzyme.
Answer:
Proteolytic.

Question 4.
The anterior part of the windpipe is enlarged to form ______ or _____.
Answer:
Larynx or Voicebox.

Question 5.
The opening of the pulmonary artery and Aorta are guarded by three ________ valves.
Answer:
Semilunar.

Question 6.
The coelomic fluid contains _____.
Answer:
Haemoglobin.

Question 7.
The Jaws of Leeches are provided with _____.
Answer:
Papillae.

Question 8.
The pain of testes is enclosed by a sac called _____.
Answer:
Scrotal Sac.

Question 9.
Central Nervous System consists of _______ and _____.
Answer:
Brain, Spinal Chord.

Question 10.
Autonomous nervous system comprises _______ and _____.
Answer:
Sympathetic, Parasympathetic.

Question 11.
Each kidney is made up of several _____.
Answer:
Nephrons.

Question 12.
In Rabbits, as male and female sexes are separate, _____ is exhibited.
Answer:
Sexual dimorphism.

Question 13.
The other name for the voice box is _____ and the other name for the windpipe is _____.
Answer:
Larynx, Trachea.

Question 14.
The right and left ventricles are separated by _____.
Answer:
Inter Ventricular Septum

Question 15.
The external ear or ______ is situated at the top of the head in Rabbits.
Answer:
Pinnae.

IV. Choose the correct answer.

Question 1.
The largest portion of the Alimentary Canal in Leech is _____.
(a) Mouth
(b) Pharynx
(c) Oesophagus
(d) Crop.
Answer:
(d) Crop.

Question 2.
In Leech, the dense network of tiny blood vessels, containing Haemocoelic fluid is called _____.
(a) Artery
(b) Capillaries
(c) Vein
(d) Blood Vessel.
Answer:
(b) Capillaries

Question 3.
Body wall of Leech includes ______ layers.
(a) 2
(b) 3
(c) 5
(d) 4.
Answer:
(c) 5

Question 4.
In female Rabbits, these four or five structures are present on the ventral surface between the thorax and abdomen _____.
(a) Nipples or Trets
(b) Limbs
(c) Vibrissae
(d) Diaphragm.
Answer:
(a) Nipples or Trets

Question 5.
In male rabbits, this structure is found in the ventral side of Anus _____.
(a) Claws
(b) Sweat glands
(c) Penis
(d) Abdomen.
Answer:
(c) Penis

V. Read the following statements and correct them, if it is not true.

Question 1.
Dorsal surface of Leech is Orange Yellow or Red in colour.
Answer:
False.
Correct Statement: Dorsal surface of Leech is Olive green in colour.

Question 2.
The clitellum is formed on segments 9 to 11, which is meant to produce cocoon, during the breeding season.
Answer:
True.

Question 3.
The lower side of the thoracic cavity in Rabbit is the dome-shaped Sternum.
Answer:
False.
Correct Statement: The lower side of the thoracic cavity in Rabbit is the dome-shaped Diaphragm.

Question 4.
Mammary glands are the modified glands of the skin.
Answer:
True.

Question 5.
The supra pharyngeal ganglion lies below the Pharynx and is formed by the fusion of four pairs of Ganglia.
Answer:
False.
Correct Statement: The sub pharyngeal ganglion lies below the Pharynx and is formed by the fusion of four pairs of Ganglia.

VI. Find the Odd One Out.

Question 1.
Cuticle, Epidermis, Sphincters, Muscular layer.
Answer:
Sphincters.

Question 2.
Undigested food, Genital pore, Rectum, Anus.
Answer:
Genital pore.

Question 3.
Gregarious, Thorax, Abdomen, Pinnae.
Answer:
Gregarious.

Question 4.
Nephrons, Nitrogenous waste, Scrotal Sac, Urea.
Answer:
Scrotal Sac.

Question 5.
Teats or Nipples, Hairs, Claws, Nails
Answer:
Teats or Nipples.

V. Answer the following in a word or sentence.

Question 1.
Where is mouth located in Leech?
Answer:
The mouth is located in the middle of the Anterior Sucker.

Question 2.
List out the structures derived from the skin of rabbit.
Answer:
The structures derived from the skin of rabbit are hair, claws, nails and glands like sweat glands, sebaceous glands and mammary glands.

Question 3.
What is Epiglottis?
Answer:
Epiglottis is a flap, in the neck, which prevents the entry of food into the trachea through the glottis.

Question 4.
The excretory system of rabbit is called a urinogenital system?
Answer:
The Urinogenital system of rabbit comprises the urinary system and genital or reproductive system. So they are called urinogenital system.

Question 5.
Where are Annular and Segmental receptors located?
Answer:
Annular receptors are located in each annulus and Segmental receptors are located on the first annulus of each segment.

Question 6.
What are Vibrissae?
Answer:
In Rabbit, from each side of upper lip tactile hairs or Whiskers present, which are called Vibrissae.

Question 7.
What is the scientific name of rabbit?
Answer:
The scientific name of rabbit is Oryctolagus cuniculus.

Question 8.
Which glands regulate the body temperature of Rabbits?
Answer:
The Sweat glands and Sebaceous glands embedded in the skin regulate the body temperature of Rabbits.

VI. Answer Briefly.

Question 1.
Explain the Excretory System of Leech briefly.
Answer:
In Leech, excretion takes place by segmentally arranged paired tubules called Nephridia. There are 17 pairs of Nephridia, which open out by nephridiopores from 6th to 22nd segments.

Question 2.
Explain the types of coelom in leech.
Answer:
Leech contains special longitudinal canals called haemocoelic canals filled with a blood like fluid called haemocolic fluid. This type of coelom is called Haemocoel.

Question 3.
What are Receptors? Write a short note on Receptors.
Answer:
Sensory projections are called Receptors on the dorsal side, there are five pairs of eyes on the first five segments. Annular receptors are located in each Annulus and segmental receptors are located on the first annulus of each segment.

Question 4.
What is hermaphrodite?
Answer:
If the male and female reproductive organs are present in the same animals, ‘ then the organism is said to hermaphrodite.

Question 5.
Name three accessory glands, which involve in the male reproductive system of Rabbit.
Answer:

• Prostate gland
• Cowper’s gland
• Perineal gland.

Their secretion involved in Reproduction.

Question 6.
Name the following:
(a) Three membranes, which cover the Brain in Rabbit
(b) Division of brain
Answer:
(a) Three membranes, which cover the Brain in Rabbit

• Dura mater
• Inner Piameter
• Arachnoid membrane.

(b) Division of brain

1. Forebrain (Prosencephalon)
2. Midbrain (Mesencephalon)
3. Hindbrain (Rhombencephalon).

Question 7.
Write a note on Medicinal value of Leech.
Answer:

1. Leeches breaks up blood clots.
2. Increases blood circulation
3. Leeches are used to treat cardio vascular diseases.
4. Saliva of leeches are used in the preparation of drugs that can treat hypertension.

Question 8.
With the Tabular Column, mention the Divisions of the body of Leech.
Answer:

Question 9.
Explain the hermaphrodite structure of leech.
Answer:
Leech is hermaphrodite because it contain both male and female genital organ in same animal.
Male Reproductive system: It is formed of: (i) Testes, (ii) Vas efferens, (Hi) Epididymis, (iv) Ejaculatory ducts and atrium.

There are eleven pairs of testes, one pair in each segment from 12 to 22 segments. They are in the form of spherical sacs called testes sacs. From each testes arises a short duct called vas efferens, which join with the vas deferens. The vas deferens becomes convoluted to form the epididymis or sperm vesicle, to store spermatozoa.

The epididymis leads to a short duct called ejaculatory duct. The ejaculatory ducts on both sides join to form the genital atrium. The atrium consists of two regions, the coiled prostate glands and the penial sac consisting of penis that opens through the male genital pore.

Female Reproductive system : It consists of ovaries, oviducts and vagina. There is a single pair of ovary in the 11th segment on the ventral side. Each ovary is a coiled ribbon-shaped structure. The ova are budded off from the ovary. From each ovary runs a short oviduct. The oviducts of the two sides joins together, to form a common oviduct. The common oviduct opens into a pear-shaped vagina which lies mid-ventrally in the posterior part of the 11th segment.

VIII. Label the following diagrams.

Question 1.
Draw and label the female Reproductive System of Rabbit.
Answer:

Question 2.
Draw and label the Digestive System of Leech.
Answer:

Question 3.
Draw and label the Brain of the rabbit.
Answer:

Question 4.
Draw and label the Lung of Rabbit.
Answer:

IX. Answer the following in Detail.

Question 1.
Explain the Digestive System of Rabbit with a neat labelled diagram.
Answer:
Teeth are hard, bone – like structures used to cut, tear and grind the food materials. The digestive system includes the Alimentary canal and the associated digestive glands. The alimentary canal consists of a mouth, buccal cavity, pharynx, oesophagus, stomach, small intestine and anus. The mouth is a transverse slit, bounded by upper and lower lips.

It leads to a buccal cavity muscular tongue is at the floor of the buccal cavity. Jaws bear teeth. Buccal cavity leads into Oesophagus. Oesophagus opens into the stomach followed by the small intestine. The thin-walled Sac, called Caecum, present at the Junction of the small intestine and large intestine, contains bacteria, which helps in digestion of cellulose.

The digestive glands such as salivary glands, gastric glands, Liver, Pancreas and intestinal glands secrete digestive juices, which help in the digestion of food in the Alimentary canal. After digestion, from the small intestine, the undigested food and wastes enter into the large intestine, which has colon and rectum. The rectum opens outside by the Anus.

Question 2.
Explain the male and the female Reproductive system of Leech with a neat labelled diagram.
Answer:
Leech is hermaphrodite, both the male and female reproductive organs are present in the same.
Male Reproductive system: There are eleven pairs of testes, one pair in each segment from 12 to 22 segments. They are in the form of Spherical Sacs called testes sacs. From each testis arises a short duct called Vas efferents, which join with the Vas deferens.

The Vas deferens becomes convoluted to form the epididymis or sperm vesicle, to store spermatozoa. The epididymis leads to a short duct called the ejaculatory duct. The ejaculatory duct on both sides joins to form Genital Atrium. The Atrium consists of prostate glands and the Penial sac, with Penis, that opens through the genital pore.

Female Reproductive system: It consists of Ovaries, Oviducts and Vagina. Each ovary is coiled, ribbon-shaped, and the single pair of the ovary is in the 11th segment. The Ova are budded off from the ovary. From each ovary runs a short oviduct, and the oviducts of two sides join together, to form a common Oviduct, which opens into a pear-shaped Vagina. Vagina lies mid ventrally in the posterior part of 11th segment.

Internal fertilisation takes place. This is followed by cocoon formation. The cocoon or egg case is formed around 9th, 10th and 11th segments. Development directs, proceeds in a cocoon, which contain one to 24 embryos. The emerging young leech resembles the adult Leech.

Question 3.
Explain the Circulatory system of Rabbit with a neat labelled diagram of a Heart.
(i) Observe the external morphology of leech specimen in your biology laboratory.
(ii) Can you find leeches in your locality?
(iii) In which geographical areas are leeches found more predominantly in India?
Answer:
The Circulatory system consists of blood, blood vessels and Heart. The Heart is pear-shaped, lies in between the lungs, in the thoracic cavity. The heart is covered by a double-layered Pericardium. The Heart is four-chambered, with two Auricles and two ventricles. The right and the left ventricles are separated by inter auricular septum. The right and the left ventricles are separated by the interventricular septum. The right auricle opens into the right ventricle, by right Auriculoventricular aperture, guarded by a tricuspid valve. The left auricle opens into the left ventricle by left Auriculoventricular aperture, guarded by a bicuspid valve or Mitral valve. The opening of the Pulmonary Artery and Aorta are guarded by three semilunar valves.

The right Auricle receives deoxygenated blood through two Precaval (superior vena cava) and one postcaval (Inferior vena cava) veins from all parts of the body. The left auricle receives oxygenated blood from the pulmonary veins from the lungs.
From the right ventricle arises Pulmonary trunk, which carries the deoxygenated blood to the lungs and from the left ventricle arises the systemic arch (aorta) which supplies Oxygenated blood to all parts of the body.

(i) Observe the external morphology of Leech in the preserved bottle specimens in the Biology Laboratory.
(ii) No, Not in our areas. The Leeches are common in hilly areas
(iii) Most Leech species are found in shallow, slow-moving freshwater and moist soil on land.

IX. Higher Order Thinking Skills (HOTS) Questions

Question 1.
How were Leeches used in Ancient times?
Answer:
Leeches were used in medicines from Ancient times. Until the 19th century, Leeches were used to draw blood from patients.

Question 2.
How are Leeches used in modern times?
Answer:
In modem times, Leeches find medical use in the treatment of Joint diseases such as Epicondylitis and Osteoarthritis, Extremity vein diseases and Microsurgery. Hirudin is a valuable dmg for some blood clotting disorders. There are doctors who will use Leeches to treat muscle cramps.

Question 3.
What is the benefit of Rabbit?
Answer:
Rabbit meat is white meat of high quality, easily digestible with low fat, low cholesterol and high protein compared to most other meats. It is an excellent source of vitamins B₃ and B₁₂, minerals, and trace elements (Phosphorus, Potassium and Selenium) Rabbit meat is an excellent balance of fatty acids, Omega 3 than chicken or pork.

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You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.3

9th Maths Algebra Exercise 3.3 Question 1.
Check whether p(x) is a multiple of g(x) or not.
p(x) = x³ – 5x² + 4x – 3, g(x) = x – 2
Solution:
p(x) = x³ – 5x² + 4x – 3; g(x) = x – 2
Let g(x) = 0
x – 2 = 0
x = 2
p(2) = 2³ – 5(2²) + 4(2) – 3
= 8 – 5 × 4 + 8 – 3 = 8 – 20 + 5 = -7 ≠ 0
⇒ p(x) is not a multiple of g(x)

9th Maths Exercise 3.3 Question 2.
By remainder theorem, find the remainder when, p(x) is divided by g(x) where,
(i) p(x) = x³ – 2x² – 4x – 1; g(x) = x + 1
(ii) p(x) = 4x³ – 12x² + 14x – 3; g(x) = 2x – 1
(iii) p(x) = x³ – 3x² + 4x + 50; g(x) = x – 3
Solution:
(i) p(x) = x³ – 2x² – 4x – 1; g(x) = x + 1
Let g(x) = x + 1
x + 1 = 0
x = -1
P(-1) = (-1)³ – 2(-1)² – 4(-1) – 1
= -1 -2 × 1 + 4 – 1
= -4 + 4 = 0
∴ Remainder = 0.

(ii) p(x) = 4x³ – 12x² + 14x – 3; g(x) = 2x – 1

(iii) p(x) = x³ – 3x² + 4x + 50 ; g(x) = x – 3
Let g(x) = x – 3
x – 3 = 0
x = 3
p(3) = 3³ – 3(3²) + 4(3) + 50
= 27 – 27 + 12 + 50
= 62
∴ Remainder = 62.

9th Maths 3.3 Question 3.
Find the remainder when 3x³ – 4x² + 7x – 5 is divided by (x + 3)
Solution:
(3x³ – 4x² + 7x – 5) + (x + 3)

The remainder is -143.

9th Class Maths Exercise 3.3 Solution Question 4.
What is the remainder when x²⁰¹⁸ + 2018 is divided by x – 1.
Solution:
x²⁰¹⁸ + 2018 is divided by x – 1
Let g(x) = x – 1 = 0
x = 1
p(x) = x²⁰¹⁸ + 2018
p(1)= 1²⁰¹⁸ + 2018
= 1 + 2018 = 2019

Samacheer Kalvi 9th Maths Chapter 3 Question 5.
For what value of k is the polynomial p(x) = 2x³ – kx² + 3x + 10 exactly divisible by (x – 2).
Solution:
Let g(x) = x – 2 = 0
x = 2
Since p(x) is exactly divisible by (x – 2)
p(2) = 2(2³) – k(2²) + 3(2)+ 10
= 16 – 4k + 6 + 10
= 32 – 4k = 0
= -k = -32
k= \(\frac{32}{4}\) = 8.

Ex 3.3 Class 9 Samacheer Question 6.
If two polynomials 2x³ + ax² + 4x – 12 and x³ + x² – 2x + a leave the same remainder when divided by (x – 3), find the value of a. and also find the remainder.
Solution:
Let f(x) = 2x³ + ax² + 4x – 12 and g(x) = x³ + x² – 2x + a
When f(x) is divided by x – 3, the remainder is f(3).
Now f(3) = 2(3)³ + a(3)² + 4(3) – 12 = 54 + 9a + 12 – 12
f(3) = 9a + 54 …………. (1)
When g(x) is divided by x – 3, the remainder is g(3).
Now g(3) = 3³ + 3² – 2(3) + a = 27 + 9 – 6 + a
g(3) = a + 30 ……….. (2)
Since, the remainder’s are same (1) = (2)
Given that f(3) = g(3)
That is 9a + 54 = a + 30
9a – a = 30 – 54 ⇒ 8a = -24 ∴ a = -3
Substituting a = -3 in f(3), we get
f(3) = 9(-3) + 54 = -27 + 54
f(3) = 27
∴ The remainder is 27.

9th Class Math 3.3 Exercise Solution Question 7.
Determine whether (x – 1) is a factor of the following polynomials:
(i) x³ + 5x² – 10x + 4
(ii) x⁴ + 5x² – 5x + 1
Solution:
(i) Let P(x) = x³ + 5x² – 10x + 4
By factor theorem (x – 1) is a factor of P(x), if P(1) = 0
P(1) = 1³ + 5(1²) – 10(1) + 4 = 1 + 5 – 10 + 4
P(1) = 0
∴ (x – 1) is a factor of x³ + 5x² – 10x + 4

(ii) Let P(x) = x⁴ + 5x² – 5x + 1
By/actor theorem, (x – 1) is a factor of P(x), if P( 1) = 0
P(1) = 1⁴ + 5 (1²) – 5(1) + 1 = 1 + 5 – 5 + 1 = 2 ≠ 0
∴ (x – 1) is not a factor of x⁴ + 5x² – 5x + 1

Class 9th Maths Exercise 3.3 Question 8.
Using factor theorem, show that (x – 5) is a factor of the polynomial 2x³ – 5x² – 28x + 15
Solution:
LetP(x) = 2x³ – 5x² – 28x + 15
By factor theorem, (x – 5) is a factor of P(x), if P(5) = 0
P(5) = 2 (5)² – 5 (5)² – 28 (5) + 15
= 2 × 125 – 5 × 25 – 140 + 15
= 250 – 125 – 140 + 15 = 265 – 265 = 0
∴ (x – 5) is a factor of 2x³ – 5x² – 28x + 15

Class 9 Maths Exercise 3.3 Solutions Question 9.
Determine the value of m, if (x + 3) is a factor of x³ – 3x² – mx + 24.
Solution:
Let P(x) = x³ – 3x² – mx + 24
By using factor theorem,
(x + 3) is a factor of P(x), then P (-3) = 0
P(-3) = (-3)³ -3 (-3)² – m (-3) + 24 = 0
⇒ -27 – 3 × 9 + 3m + 24 = 0 ⇒ 3m = 54 – 24
⇒ m = \(\frac{30}{3}\) = 10

Exercise 3.3 Class 9 Maths Question 10.
If both (x – 2) and (x – \(\frac{1}{2}\)) are the factors of ax² + 5x + b, then show that a = b.
Solution:
Let P(x) = ax² + 5x + b
(x – 2) is a factor of P(x), if P(2) = 0
P(2) = a(2)² + 5(2) + b = 0
4a + 10 + b = 0
4 a + b = – 10 …………….. (1)
(x – \(\frac{1}{2}\)) is a factor of P(x), P(\(\frac{1}{2}\)) = 0

2a + 8b = -20
a + 4b = – 10 ……………… (2)
From (1) and (2)
4a + b = – 10 …………. (1)
a + 4b = – 10 ………… (2)
(1) and (2) ⇒ 4a + b = a + 4b
3a = 3b
∴ a = b. Hence it is proved.

Ex 3.3 Class 9 Maths Solutions Question 11.
If (x – 1)divides the polynomial kx³ – 2x² + 25x – 26 without remainder, then find the value of k.
Solution:
Let P(x) = kx³ – 2x² + 25x – 26
By factor theorem, (x – 1) divides P(x) without remainder, P (1) = 0
P(1) = k(1)³ -2 (1)² + 25 (1) – 26 = 0
k – 2 + 25 – 26 = 0
k – 3 =0
k = 3

Class 9th Math Exercise 3.3 Question 12.
Check if (x + 2) and (x – 4) are the sides of a rectangle whose area is x² – 2x – 8 by using factor theorem.
Solution:
Let P(x) = x² – 2x² – 8
By using factor theorem,(x + 2) is a factor of P(x), if P (-2) = 0
P(-2) = (-2)² – 2 (-2) – 8 = 4 + 4 – 8 = 0
and also (x -4) is a factor of P (x), if P (4) = 0
p (4) = 4² – 2(4) – 8 = 16 – 8 – 8 = 0
∴ (x + 2), (x – 4) are the sides of a rectangle whose area is x² – 2x – 8.