Laxmi

Students can Download Maths Chapter 1 Life Mathematics Ex 1.4 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.4

Miscellaneous Practise Problems

Question 1.
Nanda’s marks in 3 Math tests Tl, T2 and T3 were 38 out of 40,27 out of 30 and 48 out of 50. In which test did he do well? Find his overall percentage in all the 3 tests.
Solution:
Nanda’s marks are as follows
In Test 1 → T₁ \(\frac{38}{40}\)
Test 2 → T₂ \(\frac{27}{30}\)
Test 3 → T₃ \(\frac{48}{50}\)
For percentage, multiply by 100
T₁ = \(\frac{38}{40}\) x 100 = 95%
T₂ = \(\frac{27}{30}\) x 100 = 90%
T₃ = \(\frac{48}{50}\) x 100 = 96%
Hence, he has scored highest percentage in test 3
∴ He is done well in T3
Overall percentage is the average of the 3 percentages.
i.e \(\frac{90+95+96}{3}\) = \(\frac{281}{50}\) = 93\(\frac{2}{3}\)%

Question 2.
Sultana bought the following things from a general store. Calculate the total bill amount to be paid by her.
(i) Medicines costing ₹ 800 with GST @ 5% ………..

(ii) Cosmetics costing ₹ 650 with GST @ 12% ………….

(iii) Cereals costing ₹ 900 with GST @ 0% …………

(iv) Sunglass costing ₹ 1750 with GST @ 18 % ………..

(v) Air Conditioner costing ₹ 28500 with GST @ 28% ………..

Solution:

(i) Medicine : bill amount is 800(1 + \(\frac{5}{100}\)) = 800 x \(\frac{105}{100}\) = 840
(ii) Cosmetics: Bill amount is 650(1 + \(\frac{12}{100}\)) = 650 x \(\frac{112}{100}\) = 728
(iii) Cereals : Bill amount is 900(1 + \(\frac{0}{100}\)) = 900
(iv) Sunglass: bill amount is 1750(1 + \(\frac{18}{100}\)) = 1750 x \(\frac{118}{100}\) = 2065
(v) AC : Bill amount is 28500(1 + \(\frac{28}{100}\)) = 28500 x \(\frac{128}{100}\) = 36480
∴ Total bill amount = 840 + 728 + 900 + 2065 + 36480
= ₹ 41,013 (total bill amount)

Question 3.
P’s income is 25% more than that of Q. By what percentage is Q’s income less than P’s?
Solution:
Let Q’s income be 100.
P’s income is 25% more than that of Q
∴ P’s income = 100 + \(\frac{25}{100}\) x 100 = 125
Q’s income is 25 less than that of P
In percentage terms, Q’s income is less than P’s with respect to P’s income is
\(\frac{P-Q}{P}\) x 100 = \(\frac{125-100}{125}\) x 100 = \(\frac{25}{100}\) x 100 = 20%

Question 4.
Gopi sold a laptop at 12% gain. If it had been sold for ₹ 1200 more, the gain would have been 20%. Find the cost price of the laptop.

Solution:
Let the cost price of the laptop be ‘x’
Gain = 12%

If the selling price was 1200 more
i.e \(\frac{112}{100}\)x +1200, the gain is 20%
i.e new selling price = x(1 + \(\frac{2}{100}\))
= \(\frac{112}{100}\)x + 1200 = x(100 + \(\frac{20}{100}\)) = \(\frac{112}{100}\)x
∴ 1200 = \(\frac{120}{100}\)x = \(\frac{112}{100}\)x = \(\frac{8}{100}\)x
∴ x = \(\frac{120×100}{8}\) = 15000
Cost price of the laptop is ₹ 15000

Question 5.
Vaidegi sold two sarees for ₹ 2200 each. On one she gains 10% and on the other she loses 12%. Calculate her gain or loss percentage in the sales.

Solution:
Saree 1:
The selling price is ₹ 2200, let cost price be CP₁, gain is 10%
Cost price ₹ Using the formula

Saree 2:
The selling price is 2200, let cost price be CP₂, loss is given as 12%. We need to find CP₂ using the formula as before,

Cost price of both together is CP₁ + CP₂
= 2000 + 2500 = 4500 ……(1)
Selling price of both together is 2 x 2200 = 4400 …….(2)
Since net selling price is less than net cost price, there is a loss.
loss% = \(\frac{loss}{cost price}\) x 100
Loss = Net cost price – Net selling price
(1) – (2) = 4500 – 4400 = 100
∴ loss% = \(\frac{100}{4500}\) x 100 = \(\frac{100}{45}\) = \(\frac{20}{9}\) = 2\(\frac{2}{9}\)%
= 2\(\frac{2}{9}\)% loss

Question 6.
A sum of money becomes ₹ 18000 in 2 years and ₹ 40500 in 4 years on compound interest. Find the sum.
Solution:
Let the sum of money be ‘P’
Given that sum ‘P’ becomes 18000 in 2yrs.
i.e Amount (A) = P(1 + \(\frac{r}{100}\))ⁿ
18000 = P(1 + \(\frac{r}{100}\))²
(1 + \(\frac{r}{100}\))² = \(\frac{18000}{P}\) ⇒ (1 + \(\frac{r}{100}\))⁴ = (\(\frac{18000}{P}\))² ……(1)
Also given that sum ‘P’ becomes46500 in 4 yrs.
i.e Amount (A) = P(1 + \(\frac{r}{100}\))ⁿ
40500 = P(1 + \(\frac{r}{100}\))⁴
Substituting (1) in (2), we get

Question 7.
Find the difference in the compound interest on ₹ 62500 for 1\(\frac{1}{2}\) years at 8% p.a compounded annually and when compounded half-yearly.
Solution:
Case 1:
P = ₹ 62500
n = 1\(\frac{1}{2}\)yrs. (a\(\frac{b}{c}\)) formula
r = 8% Compound annully
CI = A – P

CI = A – P = 70200 – 62500 = 7700 ……(1)
CAse 2:
P = ₹ 62500
n = 1\(\frac{1}{2}\)yrs.
r = 8% p.a when compound half yearly
r = 4% compound half yearly

= 70304 – 62500 = ₹ 7804
Difference between case 1 & case 2 is (2) – (1)
∴ (2) – (1) = 7804 – 7700 = ₹ 104

Challenging Problems

Question 8.
If the first number is 20% less than the second number and the second number is 25% more than 100, then find the first number.
Solution:
Second number is 25% more than 100
∴ 2nd number is 100 + \(\frac{25}{100}\) x 100= 125
First number is 20% less than second no.

1st number is 100

Question 9.
A shopkeeper gives two successive discounts on an article whose marked price is ? 180 and selling price is ? 108. Find the first discount percent if the second discount is 25%.
Solution:
Marked price is given as ? 180
Let 1^st discount be d₁% = ? (to find)
2nd discount be d₂% = 25%
Selling price is 108 (given)
Price after 1^st discount = 180(1 – \(\frac { d_{ 1 } }{ 100 } \)) = P₁
Price after 2^nd discount = P₁(1 – \(\frac { d_{ 2 } }{ 100 } \)) = 108
Substituting for P₁ from (1), we get

∴ 1 – \(\frac { d_{ 1 } }{ 100 } \) = \(\frac{4}{5}\)
∴ \(\frac { d_{ 1 } }{ 100 } \) = 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\)
∴ d₁ = \(\frac{1}{5}\) x 100 = 20%

Question 10.
A man bought an article on 30% discount and sold it at 40% more than the marked price. Find the profit made by him.
Solution:
Let marked price be ‘P’
Discounted price = P(1 – \(\frac{d}{100}\)) where d is the discount %
∴ Discounted price = P(1 – \(\frac{30}{100}\)) = \(\frac{70}{100}\)P → this is the cost price.
Selling price = 40% more than marked price

Question 11.
Find the rate of compound interest at which a principal becomes 1.69 times itself in 2 years.
Solution:
Let principal be ‘P’
Amount is given to be 1.69 times principal
i.e 1.69 P
Time period is 2yrs. = (n)
Rate of interest = r = ? (required)
Applying the formula,
Amount = Principal (1 + \(\frac{r}{100}\))ⁿ
Substituting, 1.69 P = P (1 + \(\frac{r}{100}\))²
∴ (1 + \(\frac{r}{100}\))² = \(\frac{1.69P}{P}\)
Taking square root on both sides, we get
\(\sqrt{1.69}\) = 1 + \(\frac{r}{100}\)
∴ 1 + \(\frac{r}{100}\) = 1.3
∴ \(\frac{r}{100}\) = 1.3
r = 30%
∴ rate of compound interst is 30%

Question 12.
The simple interest on a certain principal for 3 years at 10% p.a is ₹ 300. Find the compound interest accrued in 3 years.
Solution:
Let principal be ‘P’
Rate of interest is 10% p.a (r)
Time period (n) = 3 yrs.
Formula for simple interest
∴ P = \(\frac{300×100}{10×3}\) = 1000
Compound Interest = CI = A – P
Amount (A) = P (1 + \(\frac{r}{100}\))ⁿ
= 1000(1 + \(\frac{10}{100}\))³ = 1000 x (\(\frac{10o+10}{100}\))³
= 1000 x \(\frac{110}{100}\) x \(\frac{110}{100}\) x \(\frac{110}{100}\) = 1331
Compound Interest (Cl) = Amount – Principal
= 1331 – 1000 = 331
Compound Interest = ₹ 331

Students can Download Maths Chapter 2 Measurements Additional Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Additional Questions

Additional Questions and Answers

Exercise 2.1

Question 1.
A circular disc of radius 28 cm is divided into two equal parts. What is the perimeter of each semicircular shape disc? Also find the perimeter of the circular disc.
Answer:
Radius of the circular disc = 28 cm
∴ circumference of the circle = 2 π r units

= 2 × \(\frac { 22 }{ 7 } \) × 28 cm
= 2 × 22 × 4 = 44 × 4 = 176 cm
Perimeter of the circular disc = 176 cm
Now circumference of the semicircular disc = \(\frac { 1 }{ 2 } \) × (2 π r) = π r units
Perimeter of the semicircular disk = π r + r + r = \(\frac { 22 }{ 7 } \) × 28 + 28 + 28
= 22 × 4 + 28 + 28 = 88 + 28 + 28 = 144
Perimeter of each semicircular disc = 144 cm

Question 2.
A gardener wants to fence a circular garden of diameter 14m. Find the length of the rope he needs to purchase if he makes 3 rounds of fence. Also find the cost of the rope, if it costs ₹ 4 per metre.
Solution:
Diameter of the circular garden (d) = 14 m
Circumference = π d = \(\frac { 22 }{ 7 } \) × 14 = 44 m
Since the rope makes 3 rounds of fence,
length of the rope needed = 3 × circumference of the garden
= 3 × 44 m = 132 m
Cost of rope per meter = ₹ 4 × 132 = ₹ 528

Question 3.
Latha wants to put a lace on the edge of a circular table cover of diameter 3 m. Find the length of the lace required and also find the cost if one metre of the lace costs ₹ 30 (Take π = 3.15 )
Solution:
Diameter of the circular table cover = 3m
Circumference C = π d units = 3.15 × 3 m = 9.45 m
Length of the lace required = 9.45 m
Cost of lace per meter = ₹ 30
∴ Cost of 9.45m lace = ₹ 30 × 9.45 = ₹ 283.50

Exercise 2.2

Question 1.
The circumference of two circles are on the ratio 5 : 6. Find the ratio of their areas.
Solution:
Let the radii of the given circles be r₁ and r₂
Let their circumference be C₁ and C₂ respectively
C₁ = 2πr₁ and C₂ = 2πr₂

Now let the area of the given circles the A₁ and A₂

∴ The areas of two circles are in the ratio 25 : 36.

Question 2.
Find the area of the circle whose circumference is 88 cm.
Solution:
The circumference of the circle = 88 cm
2πr = 88 cm
2 × \(\frac { 22 }{ 7 } \) × r = 88 cm
r = \(\frac{88 \times 7}{2 \times 22}\) = 14 cm
r = 14 cm
Area of the circle A = πr²
= \(\frac { 22 }{ 7 } \) × 14 × 14 cm²
= 22 × 2 × 14 cm²
= 616 cm²

Question 3.
Find the cost of polishing a circular table top of diameter 16 dm if the rate of polishing is ₹ 15 per dm².
Solution:
Diameter of the circular table top = 16 dm
Radius (r) = \(\frac { 16 }{ 2 } \) = 8cm
Area of the circular table top
= πr² sq.units
= \(\frac { 22 }{ 7 } \) × 8 × 8 dm² = \(\frac { 1408 }{ 7 } \) dm²
= 201.14 dm²
Rate of the polishing per dm² = ₹ 15
∴ Rate of the polishing 201.14 dm²
= ₹ 15 × 201.14
= ₹ 3,017

Exercise 2.3

Question 1.
From a circular sheet of radius 5 cm a circle of radius 3 cm is removed. Find the area of the remaining sheet.
Solution:
Radius of the outer circle R = 5 cm
Radius of the inner circle r = 3 cm
Area of the remaining sheet = Area of the outer circle
– Area of the inner circle
= πR² – πr² sq. units = π(R² – r²) sq. units
= \(\frac { 22 }{ 7 } \) (5² – 3²) cm² = \(\frac { 22 }{ 7 } \) (5² – 3²) cm²
= \(\frac { 22 }{ 7 } \) × (5 + 3) (5 – 3) = \(\frac { 22 }{ 7 } \) × (8) (2)
= \(\frac { 352 }{ 7 } \) = 50.28 cm²

Question 2.
A picture is painted on a cardborad 8 cm long and 5 cm wide such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.
Solution:
Length of the outer rectangle L = 8 cm
Breadth of the outer rectangle B = 5 cm
Area of the outer rectangle = L × B sq. units = 8 × 5 cm² = 40 cm²
Length of the inner rectangle l = L – 2W = 8 – 2(1.5)cm = 8 – 3cm
l = 5cm
Breadth of thqnner rectangle b = B – 2W = 5 – 2(1.5) cm
= 5 – 3cm = 2cm
∴ Area of the margin = Area of the outer rectangle
– Area of the inner rectangle
= (40 – 10) cm² = 30cm²

Question 3.
A circular piece of radius 2 cm is cut from a rectangle sheet of length 5 cm and breadth 3 cm. Find the area left in the sheet.
Solution:
Radius of the portion removed r = 2 cm
Area of the circular sheet = πr² sq.units
= \(\frac { 22 }{ 7 } \) × 2 × 2 cm² = \(\frac { 88 }{ 7 } \) cm² = 12.57 cm²
Length of the rectangular sheet L = 5 cm
Breadth of the rectangular sheet B = 3 cm
Area of the rectangle = L × B sq. units = 5 × 3 cm² = 15 cm²
Area of the sheet left over = Area of the rectangle – Area of the circle
= 15 cm² – 12.57 cm² = 2.43 cm²

You can Download Samacheer Kalvi 9th 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 Science Solutions Chapter 8 Sound

Samacheer Kalvi 9th Science Sound Textbook Exercises

I. Choose the correct answer:

Question 1.
Which of the following vibrates when a musical note is produced by the cymbals in a orchestra?
(a) stretched strings
(b) stretched membranes
(c) air columns
(d) metal plates
Answer:
(a) stretched strings

Question 2.
Sound travels in air:
(a) if there is no moisture in the atmosphere.
(b) if particles of medium travel from one place to another.
(c) if both particles as well as disturbance move from one place to another.
(d) if disturbance moves.
Answer:
(b) if particles of medium travel from one place to another.

Question 3.
A musical instrument is producing continuous note. This note cannot be heard by a person having a normal hearing range. This note must then be passing through ………….
(a) wax
(b) vacuum
(c) water
(d) empty vessel
Answer:
(d) empty vessel

Question 4.
The maximum speed of vibrations which produces audible sound will be in ……………..
(a) seawater
(b) ground glass
(c) dry air
(d) Human blood
Answer:
(a) seawater

Question 5.
The sound waves travel faster
(a) in liquids
(b) in gases
(c) in solids
(d) in vacuum
Answer:
(c) in solids

II. Fill in the blanks.

1. Sound is a ………… wave and needs a material medium to travel.
2. Number of vibrations produced in one second is …………..
3. The velocity of sound in solid is …………. than the velocity of sound in air.
4. Vibration of object produces …………..
5. Loudness is proportional to the square of the …………..
6. …………… is a medical instrument used for listening to sounds produced in the body.
7. The repeated reflection that results in persistence of sound is called ……………..

Answer:

1. longitudinal
2. frequency of wave
3. faster
4. Sound
5. amplitude
6. ECG
7. reverberation

III. Match the following.

Answer:
(a) (v)
(b) (i)
(c) (iv)
(d) (ii)
(e) (iii)

IV. Answer in brief.

Question 1.
Through which medium sound travels faster, iron or water? Give reason.
Answer:
Sound travels faster through iron as solids are packed together tighter than liquids and gases.

Question 2.
Name the physical quantity whose SI unit is ‘hertz’. Define.
Answer:
The SI unit of frequency is hertz (Hz). The number of vibrations (complete waves or cycles) produced in one second is called frequency of the wave.

Question 3.
What is meant by supersonic speed?
Answer:
When the speed of any object exceeds the speed of sound in air (330 m s-1) it is said to be travelling at supersonic speed.

Question 4.
How does the sound produced by a vibrating object in a medium reach your ears?
Answer:
When an object vibrates, it forces the neighbouring particles of the medium to vibrate. These vibrating particles then force the particles adjacent to them to vibrate. In this way vibrations produced by an object are transferred till it reaches the ear.

Question 5.
You and your friend are on the moon. Will you be able to hear any sound produced by your friend?
Answer:
No, as there is no medium on moon for the sound to travel.

V. Answer in detail.

Question 1.
Describe with diagram, how compressions and rarefactions are produced.
Answer:
Sound is also a longitudinal wave. Sound can travel only when there are particles which can be compressed and rarefied. Compressions are the regions where particles are crowded together. Rarefactions are the regions of low pressure where particles are spread apart. A sound wave is an example of a longitudinal mechanical wave. Below figure represents the longitudinal nature of sound wave in the medium.

Question 2.
Verify experimentally the laws of reflection of sound.
Answer:
The laws of reflection are:

• The angle in which the sound is incident is equal to the angle in which sound is reflected.
• Direction of incident sound, direction of the reflected sound and the normal are in the same plane.

Take two metal tubes A and B. Keep one end of each tube on a metal plate as shown in figure. Place a wrist watch c at the open end of the tube A and interpose a cardboard between A and B. Now at a particular inclination of the tube B with the cardboard, ticking of the watch is clearly heard. The angle of reflection made by the tube B with the cardboard is equal to the angle of incidence made by the tube A with the cardboard.
For example, if the angle of incidence is 20° on the left side of a protractor (see figure arrangement), the angle of reflection at which we are able to hear the sound clearly will also be at 20° on the right side of the protractor.
∴ ∠ i = ∠ r    ➝ verifying Law I
We will also further observe that pipe 1, pipe 2 i.e, the incident ray, reflected ray and the normal lie on the same plane, verifying law II.

Question 3.
List the applications of sound.
Applications of ultrasound
Answer:

• Ultra sound can be used in cleaning technology. Minute foreign particles can be removed from objects placed in a liquid bath through which ultrasound is passed.
• Ultrasounds can also be use d to detect cracks and flaws in metal blocks.
• Ultrasonic waves are made to reflect from various parts of the heart and form the image of the heart. This technique is called ‘echo cardiography’.
• Ultrasound may be employed to break small ‘stones’ formed in the kidney into fine grains. These grains later get flushed out with urine.

Question 4.
Explain how does SONAR work?
Answer:
SONAR stands for Sound Navigation And Ranging. Sonar is a device that uses ultrasonic waves to measure the distance, direction and speed of underwater objects. Sonar consists of Science – 9 (Physics)
a transmitter and a detector and is installed at the bottom of boats and ships. The transmitter produces and transmits ultrasonic waves. These waves travel through water and after striking the object on the seabed, get reflected back and are sensed by the detector.

The detector converts the ultrasonic waves into electrical signals which are appropriately interpreted. The distance of the object that reflected the sound wave can be calculated by knowing the speed of sound in water and the time interval between transmission and reception of the ultrasound. Sonar technique is used to determine the depth of the sea and to locate underwater hills, valleys, submarine, icebergs etc.

VI. Numerical problem.

Question 1.
The frequency of a source of sound is 600 Hz. How many times does it vibrate in a minute?
Answer:
The number of vibrations (complete waves or cycles) produced in one second is called frequency of the wave.
Hence, in one minute 600 x 60 vibrations are produced = 36000 (Need clarification)

Question 2.
A stone is dropped from the top of a tower 750 m high into a pond of water at the base of the tower. When is the splash heard at the top?
(Given g = 10 m s^{– 2} and speed of sound = 340 ms^{– 1})
Height = 750m
h = ut + 0.5 gt²
The initial velocity is 0
750 = 0.5 × 10 × t²
t = 10sec
Speed of sound is 340m/sec
So, time taken to travel 750m upwards is,
\(\frac { 750 }{ 340 }\) = 2.20s
time taken = 10 + 2.20 = 12.2 sec.

Samacheer Kalvi 9th Science Sound In Text Problems

Question 1.
A sound wave has a frequency of 2 kHz and wavelength of 15 cm. How much time will it take to travel 1.5 km?
Solution:
v = nλ
n = 2 kHz = 2000Hz
λ = 15 cm = 0.15 m
0.15 × 2000 = 300ms^{– 1}
Time(t) = \(\frac{\text { Distance (d) }}{\text { Velocity (v) }}\)
The sound will take 5 s to travel a distance of 1.5 km.

Question 2.
What is the wavelength of a sound wave in air at 20° C with a frequency of 22 MHz?
Solution:
λ = v/n
Here, v = 344 m s^{– 1}
n = 22 MHz = 22 × 10⁶ Hz
λ = \(\frac { 344 }{ 22 }\) × 10⁶ = 15.64 × 10⁶ m = 15.64 µm.

Question 3.
A man fires a gun and hears its echo after 5 s. The man then moves 310 m towards the hill and fires his gun again. If he hears the echo after 3 s, calculate the speed of sound.
Solution:
Distance (d) = velocity (v) × time (t)
Distance travelled by sound when gun fires first time, 2d = v × 5 …………. (1)
Distance travelled by sound when gun fires second time, 2d – 620 = v × 3 …………… (2)
Rewriting equation (2) as,
2d = (v × 3) + 620 ……………. (3)
Equating (1) and (3), 5v = 3v + 620
2v = 620 .
Velocity of sound, v = 310 m s^{– 1}

Question 4.
A ship sends out ultrasound that returns from the seabed and is detected after 3.42 s.
If the speed of ultrasound through sea water is 1531m s^{– 1}, what is the distance of the seabed from the ship?
Solution:
We know, distance = speed × time
2d = speed of ultrasound × *time
2d = 1531 × 3.42
∴ d = \(\frac{5236}{2}\) = 2618 m
Thus, the distance of the seabed from the ship is 2618 m or 2.618 km.

Samacheer Kalvi 9th Science Sound Additional Questions

I. Answer the following questions.

Question 1.
“Sound needs a medium for propagation”. Justify with an experiment.
Answer:

Sound needs a material like air, water, steel, etc. for its propagation. It cannot travel through vacuum. This can be demonstrated by the Bell-Jar experiment.

An electric bell and an airtight glass jar are taken. The electric bell is suspended inside the airtight jar. The jar is connected to a vacuum pump. If the bell is made to ring, we will be able to hear the sound of the bell. Now when the jar is evacuated with the vacuum pump, the air in the jar is pumped out gradually and the sound becomes feebler and feebler. We will not hear any sound, if the air is fully removed (if the jar has vacuum).

Question 2.
Name the characteristics that describe a sound.
Answer:
A sound wave can be described completely by five characteristics viz – amplitude, frequency, time period, wavelength and velocity or speed.

Question 3.
What does the intensity of sound heard at a place depend on?
Answer:
The intensity of sound heard at a place depends on the following factors;

1. Amplitude of the source
2. Distance of the observer
3. Surface area of the source
4. Density of the medium
5. Frequency of the source.

Question 4.
Why is sound wave called longitudinal wave?
Answer:
As the vibration of the particles of the medium are along the direction of wave propagation, sound waves are longitudinal waves.

Question 5.
What is sound and how is it produced?
Answer:
Sound is a form of energy which produces the sensation of hearing. It is produced due to vibrations of different objects, eg. stretched vibrating wires and vibrating drums, etc.

Question 6.
How can two whales in the sea, hundreds of kilometers apart, communicate?
Answer:
Sound travels about 5 times faster in water than in air. Therefore, two whales in the sea, hundreds of kilometers apart can communicate easily.

Question 7.
What is a stethescope? What is the principle on which it works?
Answer:
Stethescope is a medical instrument used for listening to sounds produced in the body. In stethescopes, the sounds reach doctor’s ears by multiple reflections that happen in the connecting tube.

Question 8.
What is an ECG? How does it help in the field of medicine?
Answer:
The electrocardiogram (ECG) is one of the simplest and oldest cardiac investigations available. In ECG, the sound variations produced by heart is converted into electric signals. Thus an ECG is simply a representation of the electrical activity of the heart muscle as it changes with time. Usually it is printed on paper for easy analysis. The sum of this electrical activity, when amplified and recorded for just a few seconds is known as an ECG.

Students can Download Maths Chapter 2 Measurements Ex 2.4 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.4

Miscellaneous Practice Problems

Question 1.
A wheel of a car covers a distance of 3520 cm in 20 rotations. Find the radius of the wheel?
Solutions:
Distance covered by circular wheel in 20 rotation = 3520 cm
∴ Distance covered ini rotation = \(\frac { 3520 }{ 20 } \) cm = 176 cm
∴ Circumference of the wheel = 176 cm
∴ 2πr = 176
2 × \(\frac { -2 }{ 6 } \) × r = 176
r = \(\frac{176 \times 7}{2 \times 22}\)
r = 28 cm
Radius of the wheel = 28 cm

Question 2.
The cost of fencing a circular race course at the rate of ₹ 8 per metre is ₹2112. Find the diameter of the race course.
Solution:
Cost of fencing the circumference = ₹ 2112
Cost of fencing one meter = ₹ 8
∴ Circumference of the circle = \(\frac { 2112 }{ 8 } \) = 264 m
πd = 264 m
\(\frac { 22 }{ 7 } \) × d = 264
d = \(\frac{264 \times 7}{22}\) = 12 × 7 m = 84 m
∴ Diameter of the race cource = 84 m

Question 3.
A path 2 m long and 1 m broad is constructed around a rectangular ground of dimensions 120 m and 90 m respectively. Find the area of the path.
Solution:
Length of the rectangular ground l = 120 m
Breadth b = 90 m
Length of the path W₁ = 2m
Length of the path W₂ = 1m
Length of the ground with path L = 1 + 2 (W₂) = 120 + 2(1) m
= 120 + 2 = 122 m
Breadth of the ground with path B = l + 2(W₁) units
= 90 + 2(2) m = 90 + 4 m = 94 m
∴ Area of the path = (L × B) – (1 × b) sq. units
= (122 × 94) – (122 × 94) m² = 668 m²
∴ Area of the path = 668 m²

Question 4.
The cost of decorating the circumference of a circular lawn of a house at the rate of ₹55 per metre is ₹16940. What is the radius of the lawn?
Solution:
Cost of decorating the circumference = ₹ 16,940
Cost of decorating per meter = ₹ 55
∴ Length of the circumference = \(\frac { 16940 }{ 55 } \) m = 308 m
Circumference of the circular lawn = 308 m
2 × πr = 308 m
2 × \(\frac { 22 }{ 7 } \) × r = 308 m
r = \(\frac{308 \times 7}{2 \times 22}\)
r = 49 m
Radius of the lawn = 49 m

Question 5.
Four circles are drawn side by side in a line and enclosed by a rectangle as shown below.
If the radius of each of the circles is 3 cm, then calculate:
(i) The area of the rectangle.
(ii) The area of each circle.
(iii) The shaded area inside the rectangle.

Solution:
Given radius of a circle r = 3 cm
Diameter of the circle = 2r = 2 × 3 = 6 cm
Breadth of the rectangle = Diamter of the circle
B = 6cm
Length of the rectangle L = 4 × diameter of a circle
L = 4 × 6
L = 24cm

(i) Area of the rectangle = L × B sq. units
= 24 × 6 cm²
Area of the rectangle = 144 cm²

(ii) Area of the circle = πr² sq. units
= \(\frac { 22 }{ 7 } \) × 3 × 3 cm²
= \(\frac { 198 }{ 7 } \) cm²
= 28.28 cm²

(iii) Area of the shaded area = Area of the rectangle – Area of the 4 circles
= 144 – (4 × \(\frac { 198 }{ 7 } \)) cm² = 144 – \(\frac { 792 }{ 7 } \) cm²
= 144 – 113.14 cm² = 30.85 cm²

Challenge Problems

Question 6.
A circular path has to be constructed around a circular lawn. If the outer and inner circumferences of the path are 88 cm and 44 cm respectively, find the width and area of the path.
Solution:
Outer circumference of the circular lawn = 88 cm
2πR = 88 cm
Inner circumference of the lawn 2πr = 44 cm
2πR – 2πr = 88 – 44
2 × \(\frac { 22 }{ 7 } \) (R – r) = 44
(R – r) = \(\frac{44 \times 7}{2 \times 22}\)
Outer radius – Inner radius = 7 cm
∴ Width of the lawn = 7 cm
Also 2πR + 2πr = 88 + 44
2π (R + r) = 132
π (R + r) = \(\frac { 132 }{ 2 } \) = 66 cm
Area of the path = πR² – πr² sq. units
= π (R + r) (R – r) = 66 × 7
Area of the path = 462cm²

Question 7.
A cow is tethered with a rope of length 35 m at the centre of the rectangular field of length 76 m and breadth 60 m. Find the area of the land that the cow cannot graze?
Solution:
Length of the field l = 76 m
Breadth of the field b = 60m
Area of the field A = l × b sq. units = 76 × 60 m²
Area of the field A = 4560 m²
Length of the rope = 35m
Radius of the land that the cow can graze = 35m
Area of the land tha the cow can graze = circle of radius 35 m = πr² sq.units
π × 35 × 35 m² = \(\frac { 22 }{ 7 } \) × 35 × 35 m²
= 3850 m²
Area of the land the cow cannot graze = Area of the field – Area that the cow can graze
= 4560 – 3860 m² = 710 m²
Area of the land that the cow cannot graze = 710 m²

Question 8.
A path 5 m wide runs along the inside of the rectangular field. The length of the rectangular field is three times the breadth.of the field. If the area of the path is 500 m² then find the length and breadth of the field.
Solution:
Let the length of the rectangular field = ‘L’ m
Breadth of the rectangular field = = ‘B’ m
Area of the rectangular field = (L × B) m²
Also given length = 3 × Breadth
L = 3B
Width of the path (W) = 5m
Lenth of the inner rectangle = L – 2W = l – 2(5)
= 3B – 10m
Breadth of the inner rectangle = B – 2W
= B – 2(5)
= B – 10 m
Area of the inner rectangle = (3B – 10) (B – 10)
= 3B² – 10B – 30B + 100
Area of the path = Area of outer rectangle
– Area of inner rectangle
= (L × B) – (3B² – 10B – 30B + 100)
3B × B – (3B² – 40B + 100)
= 3B² – 3B² + 40B – 100
Area of the path = 40B – 100
Given area of the path = 500 m²
40B – 100 = 500
40B = 500 + 100 = 600
B = \(\frac { 600 }{ 40 } \)
B = 15m
Length of the field = 45 m; Breadth of the field = 15 m

Question 9.
A circular path has to be constructed around a circular ground. 1f the areas of the outer and inner circles are 1386 m2 and 616 m2 respectively, find the width and area of the path.
Solution:
Area of the outer circle = 1386 m²
πR² = 1386m²
Area of the inner circle = 616 m²
πr² = 616m²
Area of the path = Area of outer circle – Area of the inner circle
1386 m² – 616 m²
Area of the path = 770m²
Also πR² = 1386
R² = \(\frac{1386 \times 7}{22}\)
R² = 63 × 7
R² = 9 × 7 × 7
R² = 32 × 72
R = 3 × 7
Outer Radius R = 21 m
Again πr² = 616
\(\frac { 22 }{ 7 } \) × r² = 616
r² = 28 × 7
r² = 4 × 7 × 7
r² = 22 × 72
r = 2 × 7
Inner radius r = 14m
Width of the path = Outer radius – Inner radius = 21 – 14
Width of the path = 7m

Question 10.
A goat is tethered with a rope of length 45 m at the centre of the circular grass land whose radius is 52 m. Find the area of the grass land that the goat cannot graze.
Solution:
Length of the rope = 45 m = Radius of the inner circle
∴ Area of the circular area that the goat graze = πr² sq. units
= \(\frac { 22 }{ 7 } \) × 45 × 45 m² = 6364.28 m²
Radius of the gross land = 52 m
Area of the grass land = \(\frac { 22 }{ 7 } \) × 52 × 52 = 8,498.28 m²
Area that the goat cannot graze
= Area of the outer circle – Area of the inner circle
= 8498.28 – 6364.28 = 2134 m²
Area of the goat cannot grass = 2134 m²

Question 11.
A strip of 4 cm wide is cut and removed from all the sides of the rectangular cardboard with dimensions 30 cm × 20 cm. Find the area of the removed portion and area of the remaining cardboard.
Solution:
Area of the outer rectangular cardboard
= L × B sq.units = 30 × 20 cm² = 600 cm²
Width of the stip = 4 cm
Length of the inner rectangle = L – 2W
l = 30 – 2(4) = 30 – 8
l = 22cm
Breadth of the inner rectangle B = 2W = 20 – 2(4) = 20 – 8
b = 12cm
Area of the inner rectangle = l × b sq.units = 22 × 12 cm² = 264 cm²
Area of the remaining cardboard = 264 cm²
Area of the removed portion = Area of outer rectangle
– Area of the inner rectangle
= 600 – 264 cm²
Area of the removed portion = 336 cm²

Question 12.
A rectangular field is of dimension 20 m × 15 m. Two paths run parallel to the sides of the rectangle through the centre of the field. The width of the longer path is 2m and that of the shorter path is 1 m. Find (i) the area of the paths (ii) the area of the remaining portion of the field (iii) the cost of constructing the roads at the rate of ₹ 10 per sq.m.
Solution:
Length of the rectangular field L = 20 m
Breadth B = 15m
Area = L × B
20 × 15 m²
Area of outer rectangle = 300 m²

Area of inner small rectangle = \(\frac { 19 }{ 2 } \) × \(\frac { 13 }{ 2 } \) = 61.75 cm²

(i) Area of the path = Area of the outer rectangle
– Area of 4 inner small rectangles
= 300 – 4(61.75) = 300 – 247 = 53 m²
Area of the paths = 53 m²

(ii) Area of the remaining portion of the field
= Area of the outer rectangle – Area of the paths
= 300 – 53 m² = 247 m²
Area of the remaining portion = 247 m²

(iii) Cost of constructing 1 m² road = ₹10
∴ Cost of constructing 53 m² road = ₹10 × 53 = ₹530
∴ Cost of constructing road = ₹530

You can Download Samacheer Kalvi 9th 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 Science Solutions Chapter 9 Universe

Samacheer Kalvi 9th Science Universe Textbook Exercises

I. Choose the correct answer.

Question 1.
Who proposed the heliocentric model of the universe?
(a) Tycho Brahe
(b) Nicolaus Copernicus
(c) Ptolemy
(d) Archimedes
Answer:
(b) Nicolaus Copernicus

Question 2.
Which of the following is not a part of outer solar system?
(a) Mercury
(b) Saturn
(c) Uranus
(d) Neptune
Answer:
(a) Mercury

Question 3.
Ceres is a …………..
(a) Meteor
(b) Star
(c) Planet
(d) Astroid
Answer:

Question 4.
The period of revolution of planet A around the Sun is 8 times that of planet B. How many
times is the distance of planet A as great as that of planet B?
(a) 4
(b) 5
(c) 2
(d) 3
Answer:
(c) 2

Question 5.
The Big Bang occurred years ago.
(a) 13.7 billion
(b) 15 million
(c) 15 billion
(d) 20 million
Answer:
(a) 13.7 billion

II. Fill in the blanks.

1. The speed of Sun in km/s is ……………..
2. The rotational period of the Sun near its poles is ………………..
3. India’s first satellite is …………….
4. The third law of Kepler is also known as the Law of ………………
5. The number of planets in our Solar System is ………………

Answer:

1. 250
2. 36 days
3. Aryabhatta
4. Harmonies 5.8

III. True or False.

1. ISS is a proof for international cooperation – True.
2. Halley’s comet appears after nearly 67 hours -False.
   Correct statement: Halley’s comet appears after nearly every 76 years.
3. Satellites nearer to the Earth should have lesser orbital velocity – False.
   Correct statement: Nearer the object to the Earth, the faster is the required orbital velocity.
4. Mars is called the red planet – True

IV. Answer very briefly.

Question 1.
What is solar system?
Answer:
The Sun and celestial bodies which revolve around it form the solar system. It consists of large number of bodies such as planets, comets, asteroids and meteors.

Question 2.
Define orbital velocity.
Answer:
The horizontal velocity that has to be imparted to a satellite at the determined height so that it makes a circular orbit around the planet is called orbital velocity.

Question 3.
Define time period of a satellite.
Answer:
Time taken by the satellite to complete one revolution round the Earth is called time period.
Time period, T = Distance covered/Orbital velocity
T = 2πr/v

Question 4.
What is a satellite? What are the two types of satellites?
Answer:
A body moving in an orbit around a planet is called satellite. The two types of satellites are natural and artificial. Earth’s natural satellite is the moon.

Question 5.
Write a note on the inner planets.
Answer:
The planets are spaced unevenly. The first four planets are relatively close together and close to the Sun. They form the inner solar system. The four planets grouped together in the inner solar system are Mercury, Venus, Earth and Mars. They are called inner planets.
They have a surface of solid rock crust and so are called terrestrial or rocky planets. Their insides, surfaces and atmospheres are formed in a similar way and form similar pattern. Our planet, Earth can be taken as a model of the other three planets.

Question 6.
Write about comets in brief.
Answer:
Comets are lumps of dust and ice that revolve around the Sun in highly elliptical orbits. Their period of revolution is very long. When approaching the Sun, a comet vaporizes and forms a head and tail. Some of the biggest comets even seen had tails 160 million (16 crores) km long. Many comets are known to appear periodically. One such comet is Halley’s Comet, which appears after nearly every 76 years. It was last seen in 1986. It will next be seen in 2062.

Question 7.
State Kepler’s laws.
Answer:
In the early 1600s, Johannes Kepler proposed three laws of planetary motion.
First Law – The Law of Ellipses
The path of the planets about the Sun is elliptical in shape, with the center of the Sun being located at one of the foci.

Second Law – The Law of Equal Areas
An imaginary line drawn from the center of the Sun to the center of the planet will sweep out equal areas in equal intervals of time.

Third Law – The Law of Harmonies
The ratio of the squares of the periods of any two planets is equal to the ratio of the cubes of their semi major axis from the Sun.

Question 8.
What factors have made life on Earth possible?
Answer:
The Earth where we live is the only planet in the solar system which supports life. Due to its right distance from the Sun it has the right temperature, the presence of water and suitable atmosphere and a blanket of ozone. All these have made continuation of life possible on the Earth. The axis of rotation of the Earth is not perpendicular to the plane of its orbit. The tilt is responsible for the change of seasons on the Earth.

V. Answer in detail.

Question 1.
Give an account of all the planets in the solar system.
Answer:
The planets are spaced unevenly. The first four planets are relatively close together and close to the Sun. They form the inner solar system. Farther from the Sun is the outer solar system, where the planets are much more spread out. Thus the distance between Saturn and Uranus is much greater (about 20 times) than between the Earth and the Mars.

The four planets grouped together in the inner solar system are Mercury, Venus, Earth and Mars. They are called inner planets. They have a surface of solid rock crust and so are called terrestrial or rocky planets. Their insides, surfaces and atmospheres are formed in a similar way and form similar pattern. Our planet, Earth can be taken as a model of the other three planets.

The four large planets Jupiter, Saturn, Uranus and Neptune spread out in the outer solar system that slowly orbit the Sun are called outer planets. They are made of hydrogen, helium and other gases in huge amounts and have very dense atmosphere. They are known as gas giants and are called gaseous planets. The four outer planets Jupiter, Saturn, Uranus and Neptune have rings whereas the four inner planets do not have any rings. The rings are actually tiny pieces of rock covered with ice.

Question 2.
Discuss the benefits of ISS.
Answer:
International Space Station (ISS), is a large spacecraft which can house astronauts. It goes around in low Earth orbit at approximately 400 km distance. It is also a science laboratory. Its very first part was placed in orbit in 1998 and its core construction was completed by 2011. It is the largest man-made object in space which can also be seen from the Earth through the naked eye. The first human crew went to the ISS in 2000. Ever since that, it has never been unoccupied by humans. At any given instant, at least six humans will be present in the ISS. According to the current plan ISS will be operated until 2024, with a possible extension until 2028. After that, it could be deorbited, or recycled for future space stations.

The ISS is intended to act as a scientific laboratory and observatory. Its main purpose is to provide an international lab for conducting experiments in space, as the space environment is nearly impossible to reproduce here on Earth. The microgravity environment present in the ISS provides ideal conditions for doing many scientific researches especially in biology, human biology, physics, astronomy and meteorology.

Question 3.
Write a note on orbital velocity.
Answer:
Nowadays many artificial satellites are launched into the Earth’s orbit. The first artificial satellite Sputnik was launched in 1956. India launched its first satellite Aryabhatta on April 19, 1975. Artificial satellites are made to revolve in an orbit at a height of few hundred kilometres. At this altitude, the friction due to air is negligible. The satellite is carried by a rocket to the desired height and released horizontally with a high velocity, so that it remains moving in a nearly circular orbit. The horizontal velocity that has to be imparted to a satellite at the determined height so that it makes a circular orbit around the planet is called orbital velocity.

The orbital velocity of the satellite depends on its altitude above Earth. Nearer the object to the Earth, the faster is the required orbital velocity. At an altitude of 200 kilometres, the required orbital velocity is little more than 27,400 kph. That orbital speed and distance permit the satellite to make one revolution in 24 hours. Since Earth also rotates once in 24 hours, a satellite stays in a fixed position relative to a point on Earth’s surface. Because the satellite stays over the same spot all the time, this kind of orbit is called ‘geostationary’.

Orbital velocity can be calculated using the following formula.
v = \(\sqrt{\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})}}\)
where; G = Gravitational constant (6.673 × 10^{– 11}Nm²kg^{– 2})
M = Mass of the Earth (5.972 × 10²⁴kg)
R = Radius of the Earth (6371 km)
h = Height of the satellite from the surface of the Earth.

VI. Conceptual questions.

Question 1.
Why do some stars appear blue and some red?
Answer:
Stars are the fundamental building blocks of galaxies. Stars were formed when the galaxies were formed during the Big Bang. Stars produce heat, light, ultraviolet rays, x-rays, and other forms of radiation. They are largely composed of gas and plasma (a superheated state of matter).

The brightness of a star depends on their intensity and the distance from the Earth. Stars also appear to be in different colours depending on their temperature. Hot stars are white or blue, whereas cooler stars are orange or red in colour.

Question 2.
How is a satellite maintained in nearly circular orbit?
Answer:
Artificial satellites are made to revolve in an orbit at a height of few hundred kilometres. At this altitude, the friction due to air is negligible. The satellite is carried by a rocket to the desired height and released horizontally with a high velocity, so that it remains moving in a nearly circular orbit.

Question 3.
Why are some satellites called geostationary?
Answer:
Since Earth also rotates once in 24 hours, a satellite stays in a fixed position relative to a point on Eqrth’s surface. Because the satellite stays over the same spot all the time, this kind of orbit is called ‘geostationary’.

Question 4.
A man weighing 60 kg in the Earth will weigh 1680 kg in the Sun. Why?
Answer:
Mass of the man = 60Kg
w = m × g
m = 60Kg, g = 274.13m/s²
The sun’s gravtiational acceleration is 30 times more than that of the earth. So the person would weigh 16,447N on the surface of sun.

VII. Numerical problems.

Question 1.
Calculate the speed with which a satellite moves if it is at a height of 36,000 km from the Earth’s surface and has an orbital period of 24 hr (Take R = 6370 km). [Hint: Convert hr into seconds before doing calculation]
Answer:
T = 2π(R+h)/v
86400 = 2 × 3.14 × (6370 + 36000)/v
v = 6.28 × 42370
266083.6 km/sec
S = d/t
= 266083/24
= 11086.79 km/h

Question 2.
At an orbital height of 400 km, find the orbital period of the satellite.
Answer:
h = 400 × 10³m, R = 6371 × 10³m,
v = 7616 × 10³ kms^{– 1}.
Substituting the values,
T = 2π(R+h)/v
T = 6.28 × \(\frac { 6771 }{ 7616 }\)
T = 5.583 × 10³s = 5583 .
T ≈ 93 min

Activity

Question 1.
Watch the sky in the early morning. Do you see any planet? What is its name? Find out with the help of your teachers.
Answer:
The planet that can be observed in the early morning is ‘Venus’.

Question 2.
Prepare a list of Indian satellites from Aryabhatta to the latest along with their purposes.
Answer:

Samacheer Kalvi 9th Science Universe In Text Problems

Question 1.
Can you calculate the orbital velocity of a satellite orbiting at an altitude of 500 km?
Data: G = 6.673 × 10^{– 11} SI units;
M = 5.972 × 10²⁴ kg; R = 6371000 m;
h = 500000 m.
Solution:
v = \(\sqrt{\frac{6.67 \times 10^{-11} \times 5.972 \times 10^{24}}{(6371000+500000)}}\)
∴ v = 7613 ms^{– 1} or 7.613 kms^{– 1}

Question 2.
At an orbital height of 500 km, find the orbital period of the satellite.
Solution:
h = 500 × 10³m, R = 6371 × 10³m,
v = 7616 × 10³ kms^{– 1}.
T = \(\frac{2 \pi(\mathrm{R}+\mathrm{h})}{\mathrm{v}}=2 \times \frac{22}{7} \times \frac{(6371+500)}{7616}\)
= 5.6677 × 10³s = 5667 s.
This is T ≈ 95 min

Samacheer Kalvi 9th Science Universe Additional Questions

I. Answer the following questions.

Question 1.
What is the composition of sun? How was it formed?
Answer:
Three quarters of the sun has hydrogen gas and one quarter has helium gas. It is over a million times as big as earth. Hydrogen atoms combine or fuse together to form helium under enormous pressure. This process is called nuclear fusion which releases enormous amount of energy as light and heat. It is this energy which makes sunshine and provide heat. The sun is situated at the centre of the solar system.

At the time of the Big Bang, hydrogen gas condensed to form huge clouds, which later concentrated and found numerous galaxies. Some hydrogen gas was left free in our galaxy. With time some changes occured due to which this free floating hydrogen gas concentrated and paved the way for formation of sun and the solar system.

Question 2.
What is a star? What is its composition and how was it formed?
Answer:
Stars are the fundamental building blocks of galaxies. Stars were formed when the galaxies were formed during the big bang. Stars produce heat, light, ultra violet rays and other forms of radiation. They are largely composed of gas and plasma. Stars are built by hydrogen gases. Hydrogen atom fuse together to form helium atoms and in the process they produce large amount of heat.

Question 3.
How is the pole star distinguished from other stars? Explain.
Answer:
Pole star was one of the most familiar stars for travellers in earlier times when there were no instruments or devices to find directions at night. It appears stationary because it is situated in the northern direction along the axis of rotation of the Earth.

Question 4.
What is the difference between a meteor and a meteorite?
Answer:
Meteors are stony or rocky bodies scattered throughout the solar system and are very small in size. The meteors are also called shooting stars. They travel at a high speed, these pieces come closer to the Earth’s atmosphere and are attracted by the gravitational force of Earth. Some are dust particles left behind by comets and other are pieces of asteroids which have collided. These are quite small, some may be as small as a sand grain.

Meteors while travelling across the Earth’s atmosphere are fast moving and get heated up to a very high temperature by air friction. The heat produced is so high, that the meteor starts glowing and then burnt and lost. The path of the meteor appears as a streak of light in the night sky. Very large meteors are able to survive from such heat destruction and actually reach the Earth’s surface. These meteors are called meteorites.

Question 5.
What is microgravity?
Answer:
It is a condition in which people or objects appear to be weightless. The effects of microgravity can be seen when astronauts and objects float in space. ‘Micro’ means small, so microgravity refers to a condition where gravity is so small. Many things seem to act differently in microgravity. Fire bums differently. Without the pull of gravity, flames are mere sound.

Students can Download Maths Chapter 2 Measurements Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Intext Questions

Exercise 2.1

Try These (Text book Page No. 23)

Question 1.
A few real life examples of circular shapes are given below.

Can you give three more examples.
Solution:

1. One and Two rupee coins
2. Bangles
3. Mouth of Bottle

Question 2.
Find the diameter of your bicycle wheel?
Solution:
Diameter of my bicycle wheel is 700 mm

Question 3.
If the diameter of the circle is 14cm, what will be it’s radius?
Solution:
diameter d = 14 cm
radius = \(\frac { d }{ 2 } \) = \(\frac { 14 }{ 2 } \) = 7cm

Question 4.
If the radius of a bangle is 2 inches then find the diameter.
Solution:
Given radius of the bangle = 2 inches
Diameter = 2 × radius = 2 × 2 = 4 inches

Exercise 2.2

Try These (Text book Page No. 33)

Question 1.
Draw circles of different radii on a graph paper. Find the area by counting the number of squares covered by the circle. Also find the area by using the formula.
(i) Find the area of the circle, if the radius is 4.2 cm.
(ii) Find the area of the circle if the diameter is 28 cm.
Solution:
(i) Radius of the circle r = 4.2 cm
Area of the circle A = π r² sq.units
= \(\frac { 22 }{ 7 } \) × 4.2 × 4.2 cm² = 5.44 cm²

(ii) Diameter of the circle d = 28 cm
radius r = \(\frac { d }{ 2 } \) = \(\frac { 28 }{ 2 } \) = 14 cm
Area of the circle A = π r² sq.units
= \(\frac { 22 }{ 7 } \) × 14 × 14 cm² = 616 cm²

Exercise 2.3

Try These (Text book Page No. 35)

Question 1.
If the outer radius and inner radius of the circles are respectively 9 cm and 6 cm, find the width of the circular pathway.
Solution:
Radius of the outer circle R = 9 cm
Radius of the inner circle r = 6 cm
Width of the circular pathway = Radius of the outer circle
– Radius of the inner circle
= (9 – 6) cm = 3 cm
Width of the circular pathway = 3 cm

Question 2.
If the area of the circular pathway is 352 sq.cm and the outer radius is 16 cm, find the inner radius.
Solution:
Given outer radius R = 16 cm
Area of the circular pathway = πR² = πr²
Area of the circular pathway = 352 sq. cm
πR² – πr² = 352 cm²
π(R² – r²) = 352
16² – r² = \(\frac{352 \times 7}{22}\)
16² – r² = 16 × 7
16² – r² = 112
16² – 112 = r²
r² = 256 – 112
r² = 144
r = 12 cm
Inner radius r = 12 cm

Question 3.
If the area of the inner rectangular region is 15 sq.cm and the area covered by the outer rectangular region is 48 sq.cm, find the area of the rectangular pathway. Area of the outer rectangle Area of the inner rectangle Area of the rectangular pathway
Solution:
Area of the outer rectangle = 48 sq.cm
Area of the inner rectangle = 15 sq.cm
Area of the rectangular pathway = Area of the outer rectangle
– Area of the inner rectangle
= 48 – 15 = 33 cm²

Students can Download Maths Chapter 1 Life Mathematics Ex 1.3 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.3

Question 1.
Fill in the blanks

Question (i)
The compound interest on ₹ 5000 at 12% p.a for 2 years compounded annually is ………..
Answer:
₹ 1272
Hint:
Compound Interest (Cl) formula is
Cl = Amount – Principal
Amount = A (1 + \(\frac{r}{100}\))ⁿ = 5000 (1 + \(\frac{12}{100}\))²
= 5000 (1 + \(\frac{112}{100}\))² = 6272
∴ Cl = 6272 – 5000 = ₹ 1272

Question (ii)
The compound interest on ₹ 8000 at 10% p.a for 1 year, compounded half yearly is …………
Answer:
₹ 820
Hint:
Compound interest (CI) = Amount – Principal
∴ Amount = P (1 + \(\frac{r}{100}\))²ⁿ [2n as it is compounded half yearly]
r = 10% p.a, for half yearly r = 1 + \(\frac{10}{2}\) = 5
A = 8000 (1 + \(\frac{5}{100}\))^2×1 = 8000 x (\(\frac{105}{100}\))² = 8820
CI = Amount – principal
= 8820 – 8000 = ₹ 820

Question (iii)
The annual rate of growth in population of a town is 10%. If its present population is 26620, the population 3 years ago was ………..
Answer:
₹ 20,000
Hint:
Rate of growth of population r = 10%; Present population = 26620
Let population 3 years ago be x
∴ Applying the formula for population growth which is similar to compound interest,
26620 = x (1 + \(\frac{r}{100}\))³
∴ 26620 = x (1 + \(\frac{10}{100}\))³ = x (\(\frac{110}{100}\))³
∴ x = 26620 x (\(\frac{110}{100}\))³
= ₹ 20,000
The population 3 years ago was ₹ 20,000

Questions (iv)
The amount if the compound interest is calculated quarterly, is found using the formula ………….
Answer:
A = P (1 + \(\frac{r}{400}\))⁴ⁿ
Hint:
Quarterly means 4 times in a year.
∴ The formula for compound interest is
A = P (1 + \(\frac{r}{400}\))⁴ⁿ

Question (v)
The difference between the S.I and C.I for 2 years for a principal of ₹ 5000 at the rate of interest 8% p.a is …….
Answer:
₹ 32
Hint:
Difference between S.I & C.I is given by the formula
CI – SI = P (\(\frac{r}{100}\))²
Principal (P) = 5000, r = 8% p.a
∴ CI – SI = 5000 (\(\frac{8}{100}\))² = 5000 x \(\frac{8}{100}\) x \(\frac{8}{100}\) = ₹ 32

Question 2.
Say True or False

Question (i)
Depreciation value is calculated by the formula P (1 – \(\frac{r}{100}\))ⁿ
Answer:
True
Hint:
Depreciation formula is P (1 – \(\frac{r}{100}\))ⁿ

Question (ii)
If the present population of a city is P and it increases at the rate of r % p.a, then the population n years ago would be P (1 + \(\frac{r}{100}\))ⁿ.
Answer:
False
Hint:
Let the population ‘n’ yrs ago be ‘x’
Present population (P) = x × (1 + \(\frac{r}{100}\))ⁿ
x = \(\frac { P }{ (1+\frac { r }{ 100 } )^{ n } } \)

Question (iii)
The present value of a machine is ₹ 16800. It depreciates @25% p.a. Its worth after 2 years is ₹ 9450.
Answer:
True
Hint:
Present value of machine = ₹ 16800
Depreciation rate (n) = 25%
Value after 2 years = P (1 – \(\frac{r}{100}\))ⁿ = 16800 (1 – \(\frac{25}{100}\))²
= 16800 x (1 – \(\frac{1}{4}\))² = 16800 x \(\frac{3}{4}\) x \(\frac{3}{4}\) = 9450

Question (iv)
The time taken for ₹ 1000 to become ₹ 1331 @20% p.a compounded annually is 3 years.
Answer:
False
Principal money = 1000
rate of interest Amount = 20%
Amount = 1331, applying in formula we get
A = (1 + \(\frac{r}{100}\))ⁿ
1331 = 1000(1 + \(\frac{r}{100}\))ⁿ
∴ \(\frac{1331}{1000}\) = (1 – \(\frac{1}{5}\))ⁿ
\(\frac{1331}{1000}\) = (\(\frac{6}{5}\))ⁿ
∴ n ≠ 3 (False)

Question (v)
The compound interest on ₹ 16000 for 9 months @20% p.a, compounded quarterly is ₹ 2522.
Answer:
True
Hint:
Principal (P) = 16000
n = 9 months = \(\frac{9}{12}\) years
r = 20% p.a
For compounding quarterly, we have to use below formula.
Amount (A) = P x (1 + \(\frac{r}{100}\))⁴ⁿ
Since quarterly we have to divide r by 4
r = \(\frac{20}{4}\)

∴ Interest = A – P = 18522 – 16000 = 2522 (True)

Question 3.
Find the compound interest on ₹ 3200 at 2.5% p.a for 2 years, compounded annually.
Solution:
Principal (P) = ₹ 3200
r = 2.5% p.a
n = 2 years comp, annually
∴ Amount (A) = (1 + \(\frac{r}{100}\))ⁿ = (1 + \(\frac{2.5}{100}\))²
= 3200 x (1.025)² = 3362
Compound interest (Cl) = Amount – Principal = 3362 – 3200 = ₹ 162

Question 4.
Find the compound interest for 2\(\frac{1}{2}\) years on ₹ 4000 at 10% p.a if the interest is compounded yearly.
Solution:
Principal (P) = ₹ 4000
r = 10% p.a
Compounded yearly n = 2\(\frac{1}{2}\) years. Since it is of the form a\(\frac{b}{c}\) years

= 4000x 1.1 x 1.1 x 1.05 = 5082
∴ Cl = Amount – Principal = 5082 – 4000 = 1082

Question 5.
Magesh invested ₹ 5000 at 12% p.a for one year. If the interest is compounded half yearly, find the amount he gets at the end of the year.
Solution:
Principal (P) = ₹ 5000
Interest compounded half yearly
r = 12% p.a = \(\frac{12}{2}\) = 6% for half yearly
t = 1 yr.
Since compounded half yearly, the formula to be used is
Amount A = P (1 + \(\frac{r}{100}\))²ⁿ
A = 5000 (1 + \(\frac{6}{100}\))^2×1 = 5000 x (\(\frac{106}{100}\))² = ₹ 5618

Question 6.
At what time will a sum of ₹ 3000 will amount to ₹ 3993 at 10% p.a compounded annually?
Solution:
Amount A = ₹ 3993
Principal = ₹ 3000
r = 10% p.a
n = ?

Question 7.
A principal becomes ₹ 2028 in 2 years at 4% p.a compound interest. Find the Principal.
Solution:
n = 2 years
r = rate of interest = 4% p.a
Amount A = ₹ 2028
Amount (A) = P (1 + \(\frac{r}{100}\))ⁿ
2028 = P (1 + \(\frac{4}{100}\))ⁿ
2028 = P (\(\frac{r}{100}\))²
∴ P = \(\frac{2028x100x100}{104×104}\) = ₹ 1875

Question 8.
At what rate percentage p.a will ₹ 5625 amount to ₹ 6084 in 2 years at compound interest?
Solution:
Principal (P) = ₹ 5625
Amount (A) = ₹ 6084
n = 2 years
r = ?
Amount (A) = P (1 + \(\frac{r}{100}\))ⁿ [Applying in formula]
6084 = 5625 (1 + \(\frac{r}{100}\))²
(1 + \(\frac{r}{100}\))² = \(\frac{6084}{5625}\)
Taking square root on both sides, we get
1 + \(\frac{r}{100}\) = \(\frac{78}{75}\)
\(\frac{r}{100}\) = \(\frac{78}{75}\) – 1 = \(\frac{3}{75}\) = \(\frac{1}{25}\)
∴ r = \(\frac{1}{25}\) x 100 = 4%

Question 9.
In how many years will ₹ 3375 amount to ₹ 4096 at 13\(\frac{1}{3}\)% p.a where interest is compounded half-yearly?
Solution:
Principal = ₹ 3375
Amount = ₹ 4096
r = 13\(\frac{1}{3}\)% p.a = \(\frac{40}{3}\)% p.a
Compounded half yearly r = \(\frac { \frac { 40 }{ 3 } }{ 2 } \) = \(\frac{2}{3}\)
Let no. of years be n
for compounding half yearly, formula is

Question 10.
Find the C.I on ₹ 15000 for 3 years if the rates of interest is 15%, 20% and 25% for I, II and III years respectively.
Solution:
Principal (P) = ₹ 15000
rate of interest 1 (a) = 15% for year I
rate of interest 2 (b) = 20% for year II
rate of interest 3 (c) = 25% for year III
Formula for amount when rate of interest is different for different years is

Compound Interest (Cl) = A – P = 25,875 – 15,000 = 10,875
Cl = ₹ 10,875

Question 11.
The present height of a tree is 847 cm. Find its height two years ago, if it increases at 10 % p.a.

Solution:
Present height of tree = 847 cm
Present height = ‘h’
n = 2 yrs
rate of growth = 10% p.a
Applying in formula, we get

∴ Original height of tree = 70 cm

Question 12.
Find the difference between the C.I and the S.I on ₹ 5000 for 1 year at 2% p.a, if the interest is compounded half yearly.
Solution:
Principal (P) = ₹ 5000
time period (n) = 1 yr.
Rate of interest (r) = 2% p.a
for half yearly r = 1%
Difference between Cl & SI is given by the formula
CI – SI = P (\(\frac{r}{100}\))²ⁿ [for half yearly compounding]
CI – SI = P (\(\frac{1}{100}\))^2×1
= 5000 x \(\frac{1}{100}\) x \(\frac{1}{100}\) = ₹ 0.50

Question 13.
What is the difference in simple interest and compound interest on 115000 for 2 years at 6% p.a compounded annually.
Solution:
Principal (P) = ₹ 15,000
Time period (n) = 2 yrs.
Rate of interest (r) = 6% p.a compounded annually
Difference between CI and SI given by
CI – SI = P (\(\frac{r}{100}\))ⁿ = 15000 (\(\frac{6}{100}\))²
= 15000 x \(\frac{6}{100}\) x \(\frac{6}{100}\)
= ₹ 54

Question 14.
Find the rate of interest if the difference between the C.I and S.I on ₹ 8000 compounded annually for 2 years is ₹ 20.
Solution:
Principal (P) = ₹ 8000
time period (n) = 2 yrs.
rate of interest (r) = ?
Difference between Cl & SI is given by the formula
CI – SI = P (1 + \(\frac{r}{100}\))ⁿ
Difference between Cl & SI is given as 20
∴ 20 = 8000 x (\(\frac{r}{100}\))²
∴ (\(\frac{r}{100}\))² = \(\frac{20}{8000}\) = \(\frac{1}{400}\)
Taking square root on both sides
\(\frac{r}{100}\) = \(\sqrt { \frac { 1 }{ 400 } } \) = \(\frac{1}{20}\)
∴ r = \(\frac{100}{20}\)

Question 15.
Find the principal if the difference between C.I and S.I on it at 15% p.a for 3 years is ₹ 1134.
Solution:
Rate of interest (r) = 15% p.a
time period (n) = 3 years
Difference between Cl & SI is given as 1134
Principal = ? → required to find
Simple Interest SI = \(\frac{Pnr}{100}\)
Compound Interest CI = P (1 + i)ⁿ – P
Cl – SI = P [(1 + i)ⁿ – 1 – \(\frac{nr}{100}\)]

Objective Type Questions

Question 16.
The number of conversion periods, if the interest on a principal is compounded every two months is ……………
(a) 2
(b) 4
(c) 6
(d) 12
Answer:
(c) 6
Hint:
Conversion period is the time period after which the interest is added to the principal. If principal is compounded every two months then in a year, there will be 6\(\frac{12}{2}\) conversation periods.

Question 17.
The time taken for ₹ 4400 to become ₹ 4851 at 10%, compounded half yearly is
(a) 6 months
(b) 1 year
(c) 1\(\frac{1}{2}\) years
(d) 2years
Answer:
(b) 1 year
Hint:
Principal = ₹ 4400
Amount = ₹ 4851
Rate of interest = 10% p.a
for half yearly, divide by 2,
r = \(\frac{10}{2}\) = 5 %
Compounded half yearly, so the formula is
A = P (1 + \(\frac{r}{100}\))²ⁿ
Substuting in the above formula, we get

Taking square root on both sides, we get
(\(\frac{21}{20}\))²ⁿ = (\(\frac{21}{20}\))²
Equating power on both sides
∴ 2n = 2,
n = 1

Question 18.
The cost of a machine is ₹ 18000 and it depreciates at 16\(\frac{2}{3}\)% annually. Its value after 2 years will be ………..
(a) ₹ 12000
(b) ₹ 12500
(c) ₹ 15000
(d) ₹ 16500
Answer:
(b) ₹ 12500
Hint:
Cost of machine = ₹ 18000
Depreciation rate = 16\(\frac{2}{3}\)% = \(\frac{50}{3}\)% p.a
time period = 2 years
∴ As per depreciation formula,
Depriciated value = Original value (1 – \(\frac{r}{100}\))ⁿ
Substituting in above formula, we get

Question 19.
The sum which amounts to ₹ 2662 at 10% p.a in 3 years compounded yearly is ………..
(a) ₹ 2000
(b) ₹ 1800
(c) ₹ 1500
(d) ₹ 2500
Answer:
(a) ₹ 2000
Hint:
Amount = ₹ 2662
rate of interest = 10% p.a
Time period = 3 yrs. Compounded yearly
Principal (P) → required to find?
Applying formula A = P (1 + \(\frac{r}{100}\))ⁿ

Question 20.
The difference between simple and compound interest on a certain sum of money for 2 years at 2% p.a is ₹ 1 the sum of money is ……….
(a) ₹ 2000
(b) ₹ 1500
(c) ₹ 3000
(d) ₹ 2500
Answer:
(d) ₹ 2500
Difference between Cl and SI is given as Re 1
Time period (n) = 2 yrs.
Rate of interest (r) = 2% p.a
Formula for difference is
CI – SI = P x (1 + \(\frac{r}{100}\))ⁿ
Substituting the values in above formula, we get
1 = P x (\(\frac{2}{100}\))²
∴ P = 1 x (\(\frac{100}{2}\))²
= 1 x (50)² = ₹ 2500

Students can Download Maths Chapter 2 Measurements Ex 2.3 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.3

Question 1.
Find the area of a circular pathway whose outer radius is 32 cm and inner radius is 18 cm.
Solution:
Radius of the outer circle R = 32 cm
Radius of the inner circle r = 18 cm
Area of the circular pathway = π (R² – r²) sq. units = \(\frac { 22 }{ 7 } \) (32² – 18²) cm²
= \(\frac { 22 }{ 7 } \) × (32 + 18) × (32 – 18) cm²
= \(\frac { 22 }{ 7 } \) × 50 × 14 cm² = 2,200 cm²
Area of the circular pathway = 2,200 cm²

Question 2.
There is a circular lawn of radius 28 m. A path of 7 m width is laid around the lawn. What will be the area of the path?
Solution:
Radius of the circular lawn r = 28 m
Radius of the lawn with path = 28 + 7 m = 35 m
Area of the circular path = π (R² – r²) sq. units
Area of the path = \(\frac { 22 }{ 7 } \) (35² – 28²) m² = \(\frac { 22 }{ 7 } \) × (35 + 28) (35 – 28) m²
= \(\frac { 22 }{ 7 } \) × 63 × 7 m² = 1386 m²
Area of the path = 1386 m²

Question 3.
A circular carpet whose radius is 106 cm is laid on a circular hall of radius 120 cm. Find the area of the hall uncovered by the carpet.
Solution:
Radius of the circular hall R = 120 cm
Radius of the circular carpet r = 106 cm
Area of the hall uncovered = Area of the hall – Area of the carpet
= π (R² – r²) cm²
= \(\frac { 22 }{ 7 } \) × (120² – 106²) cm²
= \(\frac { 22 }{ 7 } \) × (120 + 106) × (120 – 106) cm²
= \(\frac { 22 }{ 7 } \) × 226 × 14 cm² = 9,944 cm²
Area of the hall uncovered = 9, 944 cm²

Question 4.
A school ground is in the shape of a circle with radius 103 m. Four tracks each of 3 m wide has to be constructed inside the ground for the purpose of track events. Find the cost of constructing the track at the rate of ₹ 50 per sq.m.

Solution:
Radius of the ground R = 103 m
Width of a track W = 3 m
Width of 4 tracks = 4 × 3 = 12 m
Radius of the ground without track
r = (103 – 12)m
r = 91 m
Area of 4 tracks = Area of the ground
– Area of the ground without crack
= πR² – πr² sq.units
= π(R² – r²) sq.units
= \(\frac { 22 }{ 7 } \) [103² – 91²]
= \(\frac { 22 }{ 7 } \) [103 + 91] [103 – 91]m²
= \(\frac { 22 }{ 7 } \) × 194 × 12 = \(\frac { 51216 }{ 7 } \) = 7316.57 m²
∴ Area of 4 tracks = 7316.57 m²
Cost of constructing 7316.57 m² = ₹ 50
∴ Cost of constructing 7316.57 m² = ₹ 50 × 7316.57 = ₹ 3,65,828,57
Cost of constructing the track ₹ 3,65,828,57

Question 5.
The figure shown is the aerial view of the pathway. Find the area of the pathway.

Solution:
Area of the rectangle = (Lenght × Breadth) sq. units
Area of the outer rectangle = (L × B) sq. units
Length of the outer rectangle L = 80 m
Breadth of the outer rectangle B = 50 m
Length of the inner rectangle l = 70 m
Breadth of the inner rectangle b = 40 m
Area of the outer rectangle = 80 × 50 m² = 4000 m²
Area of the inner rectangle = l × b sq. unit = 70 × 40 m² = 2800 m²
Area of the pathway = Area of the outer rectangle
– Area of the inner rectangle
= 4000 – 2800 m² = 1200 m²
Area of the pathway = 1200 m²

Question 6.
A rectangular garden has dimensions 11 m × 8 m. A path of 2 m wide has to be constructed along its sides. Find the area of the path.
Solution:
Area of the rectangular garden L × B = 11 m × 8 m = 88 m²
Length of the inner rectangle L = L – 2 W = 11 – 2(2) = 11 – 4 = 7 m
Breadth of the inner rectangle b = B – 2W = 8 – 2(2) = 8 – 4 = 4 m
Area of the inner rectangle = l × b sq. units = 7 × 4 m² = 28 m²
Area of the path = Area of the outer rectangular garden
– Area of the inner rectangle
= 88 m² – 28m² = 60 m²
Area of the path = 60 m²

Question 7.
A picture is painted on a ceiling of a marriage hall whose length and breadth are 18 m and 7 m respectively. There is a border of 10 cm along each of its sides. Find the area of the border.
Solution:
Length of the ceiling L = 18 m
Breadth of the ceiling B = 7 m
Area of the ceiling = L × B sq. units = 18 × 7 m² = 126 m²
Width of the boarder W = 10 cm = \(\frac { 10 }{ 100 } \) m = 0.1 m
Length of the ceiling without border = L – 2W = 18 – 2(0.1) m
= 18 – 0.2 m = 17.8 m
Breadth of the ceiling without border = B – 2W = 7 – 2 (0.1) m
= 7 – 0.2 m = 6.8 m
Area of the ceiling without border = l × b sq.units
= 17.8 × 6.8 m² = 121.04 m²
∴ Area of the border = Area of the ceiling
– Area of the ceiling without border
= 126 – 121.04 m² = 4.96 m²
Area of the border = 4.96 m²

Question 8.
A canal of width 1 m is constructed all along inside the field which is 24 m long and 15 m wide. Find (i) the area of the canal (ii) the cost of constructing the canal at the rate of ₹ 12 per sq.m.
Solution:
Length of the field L = 24 m
Width (Breadth) of the field B = 15 m
(i) Area of the field = L × B sq. units = 24 × 15 m² = 360 m²
(ii) Width of the canal (W) = 1 m
Length of the field without canal (l) = L – 2(W) = 24 – 2(1) m
= 24 – 2 m = 22 m
Width of the field without canal (b) = B – 2W = 15 – 2(1) m
= 15 – 2 m = 13 m
Area of the field without canal = l × b sq. units = 22 × 13 m² = 286 m²
Area of the canal = 360 – 286 = 74 m²
Cost of constructing 1 m² canal = ₹ 12
Cost of the constructing 74 m² canal = ₹ 12 × 74 = ₹ 888

Objective Type Question

Question 9.
The formula to find the area of the circular path is
(i) π(R² – r²) sq. units
(ii) πr² sq. units
(iii) 2πr² sq. units
(iv) πr² + 2r sq. units
Answer:
(i) π(R² – r2) sq. units

Question 10.
The formula used to find the area of the rectangular path is
(i) p(R² – r²) sq. units
(ii) (L × B) – (l × b) sq. units
(iii) LB sq. units
(iv) lb sq. units
Answer:
(ii) (L × B) – (l × b) sq. units

Question 11.
The formula to find the width of the circular path is
(i) (L – l) units
(ii) (B – b) units
(iii) (R – r) units
(iv) (r – R) units
Answer:
(iii) (R – r) units

Students can Download Maths Chapter 1 Life Mathematics Ex 1.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.2

Question 1.
Fill in the blanks:

Question (i)
Loss or gain percent is always calculated on the ………
Answer:
Cost price

Question (ii)
A mobile phone is sold for ? 8400 at a gain of 20%. The cost price of the mobile phone is …………
Answer:
₹ 7000
Hint:
Let cost price of mobile be ₹ x. Given that selling price is ? 8400 and gain is 20%

Question (iii)
An article is sold for ₹ 555 at a loss of 7\(\frac{1}{2}\)% the cost price qfthe article is ……….
Answer:
₹ 600
Hint:
Given selling price is ₹ 555 & loss is 7\(\frac{1}{2}\)%
as per formula,

Question (iv)
The marked price of a mixer grinder is ₹ 4500 is sold for ₹ 4140 after discount. The rate of discount is ……..
Answer:
8%
Hint:
Marked price is X 4500. Discounted price in ₹ 4140
∴ Discount = Marked price – Discounted price = 4500 – 4140 – 360

Question (v)
The total bill amount of a shirt costing ₹ 575 and a T-shirt costing ₹ 325 with GST of 5% is ………..
Answer:
₹ 945
Hint:
Cost price of shirt = ₹ 575 (CP)
GST = 5%

= 575 x (\(\frac { (100+5) }{ 100 } \)) = 575 x \(\frac { 105 }{ 100 } \)
= ₹ 603.75
Cost price of T-shirt = ₹ 325 (CP)
GST = 5%

= 325 x (\(\frac { (100+5) }{ 100 } \))
Total bill amount = ₹ 603.75 + ₹ 341.25 = ₹ 945

Question 2.
If selling an article for ₹ 820 causes 10% loss on the selling price, find its cost price.
Solution:
Given that selling price (SP) = ₹ 820
Loss % = 10 %

Question 3.
If the profit earned on selling an article for ₹ 810 is the same as loss on selling it for ₹ 530, then find the cost price of the article.
Solution:
Case 1: Profit = Selling price (SP) – Cost price (CP)
Case 2: Loss = Cost price (CP) – Selling price (SP)
Given that profit of case 1 = loss of case 2
∴ P = 810 – CP
L = CP – 530
Since profit (P) = loss (L)
810 – CP = CP – 530
∴ 2CP = 810 + 530 = 1340 ⇒ C.P = \(\frac{1340}{2}\)
∴ CP = 670

Question 4.
Some articles are bought at 2 for ₹ 15 and sold at 3 for ₹ 25. Find the gain percentage.
Solution:
Let cost price of one article be C.P
Given that 2 are bought for ₹ 15
∴ 2 x CP = 15 ⇒ CP = \(\frac{15}{2}\)
Let selling price of one article be SP
Given that 3 are sold for ₹ 25
∴ 3 x CP = 25 ⇒ SP = \(\frac{25}{3}\)
∴ Gain = SP – CP = \(\frac{25}{3}\) – \(\frac{15}{2}\) = \(\frac{50-45}{6}\) = \(\frac{5}{6}\)

Question 5.
If the selling price of 10 rulers is the same as the cost price of 15 rulers, then find the gain percentage.
Solution:
Let cost price of one ruler be x
Given that selling price (SP) of 10 rulers
i.e., same as cost price (CP) of 15 rulers
∴ SP of 10 rulers = 15 × x = 15x
SP of 1 ruler = \(\frac{15x}{10}\) = 1.5x
∴ Gain = SP of 1 ruler – CP of ruler = 1.5x -x = 0.5x

Question 6.
By selling a speaker for ₹ 768, a man loses 20%. In order to gain 20% how much should he sell the speaker?

Solution:
Selling price (SP) of speaker = ₹ 768
Loss % = 20%
as per formula

For gain of 20%, we should now calculate the selling price

= 960 (\(\frac{100+20}{100}\)) = 960 x \(\frac{120}{100}\)
= 96 x 12 = ₹ 1152

Question 7.
A man sold two gas stoves for ₹ 8400 each. He sold one at a gain of 20% and the other at a loss of 20%. Find his gain or loss % in the whole transaction.

Solution:
Let the CP of gas stove 1 be x and gas stove 2 be y
Given that selling price (SP) for both is the same = ₹ 8400
Case i:
First gastove: Cost price (CP) = x
Selling Price (SP) = ₹ 8400
Gain % = 20%

Cost price of first gas stove = ₹ 7000

Case 2:
Second gastove: Cost price (CP) = y
Selling price (SP) = ₹ 8400
loss % = 20%

∴ Cost price of second stove = ₹ 10,500
From (1) and (2),
Total cost price = Cost of stove 1 + Cost of stove 2
= 7000+ 10500 = 17,500
Total selling price = SP of stove 1 + SP of stove 2
= 8400 + 8400 = 16,800
Now, we find that total selling price is less than total cost price, therefore it is a loss
∴ Loss = CP – SP= 17,500 – 16,800 = 700
Loss % = \(\frac{loss}{CP}\) x 100 = \(\frac{700}{17500}\) x 100 = 4%
Loss % = 4 %

Question 8.
Find the unknowns x, y and z

Solution:
(i) Book marked price = ₹ 225 discount = 8%

∴ 225 x (\(\frac { (100-8) }{ 100 } \))
= 225 x (\(\frac { 92 }{ 100 } \)) = ₹ 207

(ii) LED TV selling price = 11970 discount = 5%,

∴ 11970 = y x \(\frac { (100-d%) }{ 100 } \)
∴ y = \(\frac { 11970×100 }{ 95 } \) = 126 x 100 = ₹ 12,600

(iii) Digital clock marked price (MP) = ₹ 750, MP = ₹ 12,600
Selling price (SP) = ₹ 615, Discount = z

∴ 615 = 750 x \(\frac { (100-z) }{ 100 } \)
∴ (100 – z) = \(\frac { 615×100 }{ 700 } \)
100 – z = 82
∴ z = 100 – 82
Discouont = 18%

Question 9.
Find the total bill amount for the data below.

Solution:

For bill amount, we should apply GST on the discounted value of the items.

Total bill amount = Bill amount of School bag + Stationary + Cosmetics + Hair drier
= 532 + 252 + 1357 + 2304
= ₹ 4,445

Question 10.
A shopkeeper buys goods at \(\frac { 4 }{ 5 } \) of its marked price and sells them at \(\frac { 4 }{ 5 } \) of the marked price find his profit percentage.
Solution:
Let marked price be MP
Given that he buys good at \(\frac { 4 }{ 5 } \) of marked price
∴ CP (cost price) = \(\frac { 4 }{ 5 } \) MP
Given that selling price (SP) = \(\frac { 7 }{ 5 } \) x MP
∴ Profit = Selling price – Cost price = \(\frac { 7 }{ 5 } \) MP – \(\frac { 4 }{ 5 } \) MP = \(\frac { 3 }{ 5 } \)  MP

Question 11.
A branded AC has a marked price of ₹ 37250. There are 2 options given for the customer.

1. Selling Price is ₹ 37250 along with attractive gifts worth ₹ 3000 (or)
2. Discount of 8% but no free gifts. Which offer is better?

Marked price of AC = ₹ 37,250
Option 1:
Selling price = ₹ 37250 & gifts worth ₹ 3000
∴ Net gain for customer = ₹ 3000 as there is no discount on AC

Option 2:
Discount of 8%, but no gift

= 37250 x \(\frac { (100-8) }{ 100 } \) = 37250 x 0.92 = 34270
∴ Savings for customer = 37250 – 34270 = ₹ 2980
Therefore, the customer gets 3000 gift in option 1 where as he is able to save only ₹ 2980 in option 2. Therefore, option 1 is better.

Question 12.
If a mattress is marked for ₹ 7500 and is available at two successive discount of 10% and 20%, find the amount to be paid by the customer.
Solution:
Marked price of mattress = ₹ 7500
Discount d₁ = 10%
Discount d₂ = 20%

= 7500 x \(\frac { (100-10) }{ 100 } \) = 7500 x \(\frac { 90 }{ 100 } \) = 6750

= 6750 x \(\frac { (100-20) }{ 100 } \) = ₹ 5400

Objective Type Questions

Question 13.
A fruit vendor sells fruits for ₹ 200 gaining ₹ 40. His gain percentage is –
(a) 20%
(b) 22%
(c) 25%
(d) 16\(\frac { 2 }{ 3 } \)
Answer:
(c) 25%
Hint:
Selling price = ₹ 200
Gain = 40
∴ CP = Selling price – gain = 200 – 40 = 160
Gain % = \(\frac { Gain }{ CP } \) x 100 = \(\frac { 40 }{ 160 } \) x 100 = 25%

Question 14.
By selling a flower pot for ₹ 528, a woman gains 20%. At what price should she sell it to gain 25%?
(a) ₹ 500
(b) ₹ 550
(c) ₹ 553
(d) ₹ 573
Answer:
(b) ₹ 550
Hint:
If selling price (SP) = ₹ 528
Gain % = 20%
∴ CP = ?

∴ 528 = CP x \(\frac { 100+20 }{ 100 } \)
∴ CP = \(\frac { 528×100 }{ 120 } \)
If gain % = 25 %, Selling ?

= 440 x \(\frac { (100+gain%) }{ 100 } \) = 440 x \(\frac { 125 }{ 100 } \)
= ₹ 550

Question 15.
A man buys an article for ₹ 150 and makes overhead expenses which are 12% of the cost price. At what price must he sell it to gain 5%?
(a) ₹ 180
(b) ₹ 168
(c) ₹ 176.40
(d) ₹ 85
Answer:
(c) ₹ 176.40
Hint:
Cost price of article = ₹ 150
Over head expenses = 12% of cost price = \(\frac { 12 }{ 100 } \) x 150 = ₹ 85
∴ Effective cost of article = 150 + 18 = ₹ 168
Now, to gain 5%, he has to sell at

Question 16.
The price of a hat is ₹ 210. What was the marked price of the hat if it is bought at 16% discount?
(a) ₹ 243
(b) ₹ 176
(c) ₹ 230
(d) ₹ 250
Answer:
(d) ₹ 250
Hint:
Let marked price be MP
Discounted price = ₹ 210
Rate of discount = 16%
As per formula:

Question 17.
The single discount which is equivalent to two successive discount of 20% and 25% is –
(a) 40%
(b) 45%
(c) 5%
(d) 22.5%
Answer:
(a) 40%
Let marked price be MP, after discount 1 of 20%,

After discount 2 of 25%,

Comparing with formula, we get
∴ This is equivalent to a single discount of 40%

You can Download Samacheer Kalvi 9th 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 Science Solutions Chapter 2 Motion

Samacheer Kalvi 9th Science Motion Textbook Exercises

I. Choose the correct answer.

Question 1.
The area under velocity-time graph represents the
(a) velocity of the moving object
(b) displacement covered by the moving object
(c) speed of the moving object
(d) acceleration of the moving object
Answer:
(d) acceleration of the moving object

Question 2.
Which one of the following is most likely not a case of uniform circular motion?
(a) Motion of the Earth around the Sun
(b) Motion of a toy train on a circular track
(c) Motion of a racing car on a circular track
(d) Motion of hour’s hand on the dial of the clock
Answer:
(a) Motion of the Earth around the Sun

Question 3.
Which of the following graph represents uniform motion of a moving particle?

Answer:
(b)

Question 4.
The centrifugal force is
(a) a real force
(b) the force of reaction of centripetal force
(c) virtual force
(d) directed towards the centre of the circular path
Answer:
(c) virtual force

II. Fill in the blanks.

1. Speed is a quantity whereas velocity is a quantity.
2. The slope of the distance-time graph at any point gives
3. Negative acceleration is called
4. Area under velocity-time graph shows

Answer:

1. scalar, vector
2. speed
3. retardation (or) deceleration
4. displacement

III. True or False.

1. The motion of a city bus in a heavy traffic road is an example for uniform motion.
2. Acceleration can get negative value also.
3. Distance covered by a particle never becomes zero but displacement becomes zero.
4. The velocity-time graph of a particle falling freely under gravity would be straight line parallel to the x axis.
5. If the velocity-time graph of a particle is a straight line inclined to X-axis then its displacement – time graph will be a straight line.

Answer:

1. False
2. True
3. True
4. False
5. True

IV. Assertion and Reason Type Question.

Mark the correct choice as:
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
(c) If assertion is true but reason is false.
(d) If assertion is false but reason is true.

Question 1.
Assertion: The accelerated motion of an object may be due to change in magnitude of velocity or direction or both of them.
Reason: Acceleration can be produced only by change in magnitude of the velocity. It does not depend the direction.
Answer:
(c) If assertion is true but reason is false.

Question 2.
Assertion: The Speedometer of a car or a motor-cycle measures its average speed.
Reason: Average velocity is equal to total displacement divided by total time taken.
Answer:
(d) If assertion is false but reason is true.

Question 3.
Assertion: Displacement of a body may be zero when distance travelled by it is not zero.
Reason: The displacement is the shortest distance between initial and final position.
Answer:
(a) If both assertion and reason are true and reason is the correct explanation of assertion.

V. Match the following.

Answer:
1. (D)
2. (C)
3. (A)
4. (B)

VI. Answer briefly.

Question 1.
Define velocity.
Answer:
Velocity is the rate of change of displacement. It. is the displacement in unit time.

Question 2.
Distinguish distance and displacement.
Answer:

Question 3.
What do you mean by uniform motion?
Answer:
An object is said to be in uniform motion if it covers equal distances in equal intervals of time how so ever big or small these time intervals may be.

Question 4.
Compare Speed and Velocity.
Answer:

Question 5.
What do you understand about negative acceleration?
Answer:
If v < u, i.e. if final velocity is less than initial velocity, the velocity decreases with time and the value of acceleration is negative. It is called negative acceleration. It is also called as retardation (or) deceleration.

Question 6.
Is the uniform circular motion accelerated? Give reasons for your answer.
Answer:
When an object is moving with a constant speed along a circular path, the velocity changes due to the change in direction. Hence it is an accelerated motion.

Question 7.
What is meant by uniform circular motion? Give two examples of uniform circular motion.
Answer:
When an object moves with constant speed along a circular path, the motion is called uniform circular motion.
Ex. (i) Revolution of Earth around the Sun
(ii) Revolution of Moon around the Earth.

VII. Answer in detail.

Question 1.
Derive equations of motion by graphical method.
Answer:
An object is in motion with initial velocity u attains a final velocity v in time t due to acceleration a, with displacement S.
Let us try to derive these equations by graphical method. Equations of motion from velocity – time graph:

Graph shows the change in velocity with time for an uniformly accelerated object. The object starts from the point D in the graph with velocity u. Its velocity keeps increasing and after time t it reaches the point B on the graph.
The initial velocity of the object = u = OD = EA
The final velocity of the object = v = OC = EB
Time = t = OE = DA
Also from the graph we know that, AB = DC

For First equation of motion
By definition, acceleration = change in velocity / time
= (final velocity – initial velocity)/time
= (OC – OD) / OE = DC / OE
a = DC/t
DC = AB = at
From the graph EB = EA + AB
v = u + at ….(1)
This is first equation of motion.

For Second equation of motion
From the graph the distance covered by the object during time t is given by the area of quadrangle DOEB
s = area of the quadrangle DOEB
s = area of the rectangle DOEA + area of the triangle DAB
= (AE × OE) + (1/2 × AB × DA)
s = ut + 1/2at² ….(2)
This is second equation of motion.

For Third equation of motion
From the graph the distance covered by the object during time t is given by the area of the quadrangle DOEB. Here DOEB is a trapezium. Then
s = area of trapezium DOEB
= 1/2 × sum of length of parallel side × distance between parallel sides
= 1/2 × (OD + BE) × OE
s = 1/2 × (u + v) × t
Since a = (v – u) / t or t = (v – u)/a
Therefore = 1/2 × (v + u) × (v – u)/a .
2as = v² = u² + 2as
v² = u² + 2as ………(3)
This is third equation of motion.

Question 2.
Explain different types of motion.
Answer:
In physics, motion, can be classified as below.

1. Linear motion: Motion along a straight line.
2. Circular motion: Motion along a circular path.
3. Oscillatory motion: Repetitive to and fro motion of an object at regular interval of time. Random motion: Motion of the object which does not fall in any of the above categories.
4. Uniform and Non-uniform motion:
   • Uniform motion: Consider a car which covers 60 km in the first hour, 60 km in second horn-, and another 60 km in the third hour and so on. The car covers equal distance at equal interval of time. We can say that the motion of the car is uniform. An object is said to be in uniform motion if it covers equal distances in equal intervals of time howsoever big or small these time intervals may be.
   • Non-uniform motion: Now, consider a bus starting from one stop. It proceeds slowly when it passes through a crowded area of the road. Suppose, it manages to travel merely 100 m in 5 minutes due to heavy traffic and is able to travel about 2 km in 5 minutes when the road is clear. Hence, the motion of the bus is non-uniform i.e. it travels unequal distances in equal intervals of time.

VIII. Exercise problems.

Question 1.
A ball is gently dropped from a height of 20m. If its velocity increases uniformly at the rate of 10 ms^{– 2}. With what velocity will it strike the ground? After what time will it strike the ground?
Given: height = 20 m
acceleration = 10 ms^{– 2}
Formula: For free falling body,

1. v = gt
2. s = 1/2 gt²
   Solution:
   time taken to strike the ground
   s = 1/2 gt²
   20m = 1/2 × 10 ms ^{– 2} × t²
   t² = \(\frac{40 m}{10 m s^{-2}}\) = 4s ²
   ∴ t = 2s
3. Velocity of the ball when it strikes ground v = gt
   v = 10ms^{– 2} × 2s
   v = 20ms^{– 1}

Question 2.
An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 m and 20 s?
Given: Diameter of circular track = 200m
time to complete = 40s
Formula: Circumference of circular track = d. π m
speed = distance/time
Solution:
Circumference of the track = d.π
= 200 × 3.14 = 628m
speed = \(\frac{628 m}{40 s}\) = 15.7 ms^{– 1}
Distance covered in 2 min 20 s = speed × time
In the given time athlete covers 31/2 rounds = 2198m

The final position of athlete is as shown in figure (A – initial position, B – final position)
∴ Displacement = Distance × time
= 200 m

Question 3.
A racing car has a uniform acceleration of 4 ms^{– 2}. What distance it covers in 10 s after the start?
Given: acceleration = 4ms^{– 2}.
time = 10s and initial velocity = 0
Formula: s = ut + 1/2 at²
Solution:
distance covered s = 0 × t + 1/2 × 4 ms^{– 2} × (10 s)².
s = 1/2 × 4 ms^{– 2} × 100 s²
= 1/2 × 400m
= 200m

Solved Examples.

Question 1.
An object travels 16 m in 4 s and then another 16 m in 2 s. What is the average speed of the object?
Solution:
Total distance travelled by the object =16m + 16m = 32m
Total time taken = 4s + 2s = 6s
Average speed = \(\frac{\text { Total distance travelled }}{\text { total time taken }}\) = \(\frac{32}{6}\) = 5.33 ms^{– 1}
Therefore, the average speed of the object is 5.33 ms^{– 1}.

Question 2.
A sound is heard 5 s later than the lightning is seen in the sky on a rainy day. Find the distance of location of lightning? Given the speed of sound = 346 ms^{– 1}
Solution:
Speed = \(\frac{\text { Distance }}{\text { time }}\)
Distance = speed x time = 346 x5 = 1730 m
Thus, the distance of location of lightning = 1730 m

Question 3.
The brakes applied to a car produce an acceleration of 6 ms^{– 2} in the opposite direction to the motion. If the car takes 2 s to stop after the application of brakes, calculate the distance traveled during this time.
Solution:
We have been given a = – 6 ms^{– 2}, t = 2s and v = 0
From the equation of motion,
v = u + at
0 = u + (- 6 × 2) .
0 = u – 12 ∴ u = 12 ms^{– 1}
s = ut + 1/2 at²
= (12 × 2) + 1/2 (- 6 × 2 × 2) .
= 24 – 12 = 12m
Thus, the car will move 12 m before it stops after the application of brakes.

Question 4.
A 900 kg car moving at 10 ms^{– 1} takes a turn around a circle with a radius of 25 m.
Determine the acceleration and the net force acting upon the car.
Solution:
When the car turns around circle, it experiences centripetal acceleration a = \(\frac{v^{2}}{r}\)
a = \(\frac{(10)^{2}}{25}=\frac{100}{25}\) ∴ a = 4ms^{– 2}
Net force acting upon the car,
F = ma = 900 × 4 = 3600N

ACTIVITY

Question 1.
Look around you. You can see many things: a row of houses, large trees, small plants, flying birds, running cars and many more. List the objects which remain fixed at their position and the objects which keep on changing their position.
Answer:

1. The objects which remain fixed at their position, and do not change their position are a row of houses, large trees and small plants.
2. The objects which keep on changing their position are flying birds, running cars and buses.

Question 2.
Tabulate the distance covered by a bus in a heavy traffic road in equal intervals of time and do the same for a train which is not in an accelerated motion. From your table what do you understand?
Answer:
The bus covers unequal distance in equal intervals of time but the train covers equal distances in equal intervals of time.
(The students can do this activity by themselves)

Question 3.
Observe the motion of a car as shown in the figure and answer the following questions:

Compare the distance covered by the car through the path ABC and AC. What do you observe? Which path gives the shortest distance to reach D from A? Is it the path ABCD or the path ACD or the path AD?
Answer:

1. AB + BC
   ABC = 4m + 3m=7m
   AC = 5 m
2. AB + BC + CD
   ABCD = 4m + 3m + 4m = 11m
   ACD = AC + CD = 5m + 4m = 9m
   AD = 3 m
   The shortest distance to reach D from A will be AD, that is 3 m.

Question 4.
Take a large stone and a small eraser. Stand on the top of a table and drop them simultaneously from the same height?
What do you observe? Now, take a small eraser and a sheet of paper. Drop them simultaneously from the same height?
What do you observe? This time, take two sheets of paper having same mass and crumple one of the sheets into a ball. Now, drop the sheet and the ball from the same height. What do you observe?
Answer:
Both the stone and the eraser have reached the surface of the Earth almost at the same time in the absence of air medium (vacuum). But in air medium, due to friction, air offers resistance to the motion of free falling objects. The eraser reaches the first, the sheet of paper reaches later. The air resistance exerted on the sheet of paper is much higher than that of the eraser. The paper crumpled into a ball reaches ground first and plain sheet of paper reaches later, although they have equal mass. The air resistance offer to the plain sheet of paper is much higher than that offered to the paper ball. This is because the magnitude of air resistance depends on the area of objects exposed to air.

Question 5.
Take a piece of thread and tie a small piece of stone at one of its ends. Rotate the stone to describe a circular path with constant speed by holding the thread at the other end. Now, release the thread and let the stone go. Can you tell the direction in which the stone moves after it is released?
Answer:
The stone moves along the straight line tangential to the circular path. This is because once the stone releases, it continues to move along the direction it has been moving at that instant.

Question 6.
Take a piece of rope and tie a small stone at one end. Hold the other end of the rope and rotate it such that the stone follows a circular path.
Do you experience any pull or push in your hand?

In this activity, a pulling force that acts away from the centre is experienced. This is called as centrifugal force.

Samacheer Kalvi 9th Science Motion Additional Questions:

I. Choose the best answer.

Question 1.
The area under velocity time graph represents ………….
(a) Velocity of the moving object
(b) Displacement covered by the moving object
(c) Speed of the moving object
Answer:
(b) Displacement covered by the moving object

Question 2.
Unit of acceleration is ………………
(a) ms^{– 1}
(b) ms^{– 2}
(c) ms
(d) ms²
Answer:
(b) ms^{– 2}

Question 3.
When a body starts from rest, the acceleration of the body after 2 second in ………………. of its displacement.
(a) Half
(b) Twice
(c) Four times
(d) One fourth
Answer:
(a) Half

II. Short Answer Questions.

Question 1.
A bus travels, a distance of 20 km from Chennai central to airport in 45 minutes. What is the average speed?
Given: Distance = 20 km = 20,000 m
Time = 45 min = 2700 s
Formula: Average speed = \(\frac{\text { Total Distance }}{\text { Total Time taken }}\)
Solution:
Average speed = \(\frac{20 \mathrm{km}}{45 \mathrm{min}}=\frac{20,000 \mathrm{m}}{2700 \mathrm{s}}\)
= \(\frac{200 m}{27 s}\) = 7.4 ms ^{– 1}

Question 2.
Why did the actual speed differ from average, speed?
Answer:
Actual speed gives instantaneous speed of a body at any instant but average speed is the total distance covered by total time taken.

Question 3.
Mention the uses of velocity-time graph.
Answer:

1. Area covered under velocity – time graph gives us the displacement
2. Slope of the velocity – time graph gives us the acceleration.

Question 4.
The speed of a particle is constant. Will it have acceleration? Justify with an example.
Answer:

1. When a particle is moving with a constant speed along a straight line, it has no acceleration.
2. When a particle is moving with a constant speed along a circular path, the velocity changes due to the change in direction. Hence it has a acceleration.
   Ex. Revolution of Earth around the Sun.

Question 5.
Distinguish distance and displacement of a moving object.
Answer:

III. Answer the following Question briefly.

Question 1.
Derive the three equations of motion by graphical method.
An object is in motion with initial velocity u attains a final velocity v in time t due to acceleration a, with displacement s.
Let us try to derive these equations by graphical method. Equations of motion from velocity-time graph:

Graph shows the change in velocity with time for an uniformly accelerated object. The object starts from the point D in the graph with velocity u. Its velocity keeps increasing and after time t it reaches the point B on the graph.
The initial velocity of the object = u = OD = EA
The final velocity of the object = v = OC = EB
Time = t = OE = DA
Also from the graph we know that, AB = DC

For First equation of motion
By definition, acceleration = change in velocity / time
= (final velocity – initial velocity)/time
= (OC – OD) / OE = DC / OE
a = DC/t
DC = AB = at
From the graph EB = EA + AB
v = u + at …….(1)
This is first equation of motion.

For Second equation of motion
From the graph the distance covered by the object during time t is given by the area of quadrangle DOEB
s = area of the quad-rectangle DOEB
s = area of the rectangle DOEA + area of the triangle DAB
= (AE × OE) + (1/2 × AB × DA)
s = ut + 1/2at² ….(2)
This is second equation of motion.

For Third equation of motion
From the graph the distance covered by the object during time t is given by the area of the quadrangle DOEB. Here DOEB is a trapezium. Then
s = area of trapezium DOEB
= 1/2 × sum of length of parallel side × distance between parallel sides
= 1/2 × (OD + BE) × OE
s = 1/2 × (u + v) × t
since a = (v – u) / t or t = (v – u)/a
Therefore s = 1/2 × (v + u) × (v – u)/a
2as = v² – u²
v² = u² + 2 as ………..(3)
This is third equation of motion.

I. Choose the best answer.

Question 1.
In a 100 m race, the winner takes 10s to reach the finishing point. The average speed of the winner is ……….. ms^{– 1}.
Answer:
(b) 10

II. Choose correct statement.

Question 1.
(a) Action and reaction forces act on same object
(b) Action and reaction forces act on different objects
Both (a) and (b) are possible
Neither (a) nor (b) is correct
Answer:
(a) is wrong
(b) is correct.

III. Short Answer Questions.

Question 1.
A motorcycle travelling at 20 ms^{– 1} has an acceleration of 4 ms^{– 2}. What does it explains about the velocity of the motorcycle?
Answer:

It explains that the motorcycle travels with uniform velocity.

Question 2.
Complete the following sentences.
(a) The acceleration of the body that moves with a uniform velocity will be …………….. .
Answer:
constant

(b) A train travels from A to station B with a velocity of 100 km/h and returns from station B to station A with a velocity of 80 km/h. Its average velocity during the whole journey in ……………. and its average speed is ……………… .
Answer:
zero, 90 km/h

Question 3.
Distinguish speed and velocity.
Answer:

Question 4.
What is meant by negative acceleration?
Answer:
If v < u, i.e. if final velocity is less than initial velocity, the velocity decreases with time and the value of acceleration is negative. It is called negative acceleration. It is also called as retardation (or) deceleration.

IV. Answer the following Question.

Question 1.
A boy moves along the path ABCD. What is the total distance covered by the boy? What is his net displacement?
(a) Total distance covered by a boy = 110 m
(b) Net displacement = \(\sqrt{30^{2}+40^{2}}\)
= \(\sqrt{2500}\)
= 50 m