Samacheer Kalvi 10th Science Solutions Chapter 5 Pdf
Tamilnadu Board Samacheer Kalvi 10th Science Solutions Chapter 5: Tamilnadu State Board Solution Class 10 Science Chapter 5 – ACOUSTICS.
Samacheer Kalvi 10th Science Solutions Chapter 5: Overview
Board | Samacheer Kalvi |
Class | 10 |
Subject | Science |
Chapter | 5 |
Chapter Name | ACOUSTICS |
Samacheer Kalvi 10th Science Solutions Chapter 5 ACOUSTICS
I.) Choose the correct answer
1.) When a sound wave travels through air, the air particles
a) Vibrate along the direction of the wave motion
b) Vibrate but not in any fixed direction
c) Vibrate perpendicular to the direction of the wave motion
d) Do not vibrate
Ans:a) Vibrate along the direction of the wave motion
Explanation: Because, sound waves are the longitudinal waves.
2.) Velocity of sound in a gaseous medium is 330 ms–1. If the pressure is increased by 4 times without causing a change in the temperature, the velocity of sound in the gas is
a) 330 ms–1
b) 660 ms1
c) 156 ms–1
d) 990 ms–1
Ans:a) 330 ms–1
Explanation: Because, velocity of sound in air medium depends on the temperature only not on the pressure.
3.) The frequency, which is audible to the human ear is
a) 50 kHz
b) 20 kHz
c) 15000 kHz
d) 10000 kHz
Ans: b) 20 kHz
Explanation: Because, audible frequency range of hearing for human ear is 20Hz to 20 KHz.
4.) The velocity of sound in air at a particular temperature is 330 ms–1. What will be its value when temperature is doubled and the pressure is halved?
a) 330 ms–1
b) 165 ms–1
c) 330 × √2 ms–1
d) 320 / √ 2 ms–1
Ans:a) 330 ms–1
5.) If a sound wave travels with a frequency of 1.25 × 104 Hz at 344 ms–1, the wavelength will be
a) 27.52 m
b) 275.2 m
c) 0.02752 m
d) 2.752 m
Ans:c) 0.02752 m
Because, V = n*λ
Hence, λ = V/n = 344 / 1.25*104 = 0.02752
6.) The sound waves are reflected from an obstacle into the same medium from which they were incident. Which of the following changes?
a) Speed
b) Frequency
c) Wavelength
d) None of these
Ans: d) None of these
Explanation: Because, the medium of sound waves incident and reflected is same hence there will be no change in speed, frequency, wavelength of sound waves.
7) Velocity of sound in the atmosphere of a planet is 500 ms–1. The minimum distance between the sources of sound and the obstacle to hear the echo, should be
a) 17 m
b) 20 m
c) 25 m
d) 50 m
Ans: c) 25 m
II. Fill up the blanks
1) Rapid back and forth motion of a particle about its mean position is called _____
Ans: Rapid back and forth motion of a particle about its mean position is called vibrations.
2) If the energy in a longitudinal wave travels from south to north, the particles of the medium would be vibrating in _____
Ans: If the energy in a longitudinal wave travels from south to north, the particles of the medium would be vibrating in both north and south.
3) A whistle giving out a sound of frequency 450 Hz, approaches a stationary observer at a speed of 33 ms–1. The frequency heard by the observer is (speed of sound = 330 ms–1) ______.
Ans:
Given that, Velocity of sound V = 330ms-1
Velocity of whistle sound Vw = 33 ms-1
And n = 450 Hz
Thus, the frequency observed by observer i.e. apparent frequency will be
Thus, n’ = (V/ V-Vw)* n = (330/330 – 33)* 450 = 1.11
Thus, n’ = 500 Hz
4.) A source of sound is travelling with a velocity 40 km/h towards an observer and emits a sound of frequency 2000 Hz. If the velocity of sound is 1220 km/h, then the apparent frequency heard by the observer is ___________.
Ans:
Given that, Velocity of sound V = 1220km/h = 339ms-1
Velocity of whistle sound Vw = 40km/h = 11.1ms-1
And n = 2000Hz
Thus, the frequency observed by observer i.e. apparent frequency will be
Thus, n’ = (V/ V-Vw)* n = (339/339– 11.1)* 2000 = 1.03385*2000
Thus, n’ = 2068 Hz
III.) True or false: – (If false give the reason)
1) Sound can travel through solids, gases, liquids and even vacuum.
Ans: False
Correct statement: Sound waves requires material medium for their propagation and it cannot propagate through vacuum.
2) Waves created by Earth Quake are Infrasonic.
Ans: True
3) The velocity of sound is independent of temperature.
Ans: False
Correct statement: For ideal gas, the velocity of sound depends on its temperature and independent of pressure.
4) The Velocity of sound is high in gases than liquids.
Ans: False
Correct statement: Velocity of sound is high in liquids than gases.
IV.) Match the following
Ans:
1) Infrasonic: 10Hz
2) Echo: Ultrasonography
3) Ultrasonic: 22 KHz
4) High-pressure region: Compressions
V.) Assertion and Reason Questions Mark the correct choice as
a.) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
b.) If both the assertion and the reason are true but the reason is not the correct explanation of the assertion.
c.) Assertion is true, but the reason is false.
d.) Assertion is false, but the reason is true.
1) Assertion: The change in air pressure affects the speed of sound.
Reason: The speed of sound in a gas is proportional to the square of the pressure
Ans:d. Assertion is false, but the reason is true.
2) Assertion: Sound travels faster in solids than in gases.
Reason: Solid possess a greater density than that of gases.
Ans: b. if both the assertion and the reason are true but the reason is not the correct explanation of the assertion
VI.) Answer very briefly
1) What is a longitudinal wave?
Ans:
Longitudinal are the waves in which particle of the medium vibrates along the direction of propagation or parallel to the direction of propagation.
2) What is the audible range of frequency?
Ans:
3) What is the minimum distance needed for an echo?
Ans:
4) What will be the frequency sound having 0.20 m as its wavelength, when it travels with a speed of 331 ms–1?
Ans:
Given that, v = 331ms-1
Wavelength λ = 0.2 m
Hence, frequency n = v/ λ = 331/0.2 = 1655 Hz
Thus, the frequency of sound will be 1655 Hz.
5) Name three animals, which can hear ultrasonic vibrations.
Ans:
The creatures such as mosquito, dogs, bat, and dolphins can detect the ultrasonic waves.
VII.) Answer briefly
1) Why does sound travel faster on a rainy day than on a dry day?
Ans:
Because, as the humidity increases the speed of sound decreases. And in rainy season there is more humidity than on dry day. Hence, due to more humidity the speed of sound in rainy season is more than on dry day.
2) Why does an empty vessel produce more sound than a filled one?
Ans:In an empty vessel the multiple reflections of sound takes place in which intensity of sound is not decreased but in through filled vessel there is a loss of intensity of sound waves.
Hence, empty vessel produces more sound than filled one.
3) Air temperature in the Rajasthan desert can reach 46°C. What is the velocity of sound in air at that temperature? (V0 = 331 ms–1)
Ans:
Given that, v = 331 ms-1
Air temperature in desert T = 460 C
Thus velocity of sound in air temperature will be,
vT = (v + 0.61T)
= (331 + 0.61*46)
= 331 + 28.06
= 359.06 ms-1
Thus, the velocity of sound in air temperature will be 359.06 ms-1.
4) Explain why, the ceilings of concert halls are curved.
Ans:
- The ceilings of concert halls are curved because from a curved surface there are multiple reflections of sound takes place.
- When sound reflects from concave surface, the waves get reflected and converged and focused at a particular point. So at that point the intensity of sound will be more concentrated.
- Also, parabolic surfaces are used to for focusing sound waves at a point. Due to which many halls are designed with parabolic reflecting surfaces.
5) Mention two cases in which there is no Doppler effect in sound?
Ans:
In following conditions the Doppler Effect is not observed and the apparent frequency reaching to observer is same as the frequency of sound emitted from the source.
- When the source and listener both are at rest, there will be no Doppler effect is observed.
- When source and listener are moved in such way that the distance between them is remained constant.
- Also, when source and listener moves in mutual perpendicular directions there will be no Doppler effect is observed.
- When the source is located at the centre of the circle along which the listener is moving.
VIII.) Problem Corner
1) A sound wave has a frequency of 200 Hz and a speed of 400 ms–1 in a medium. Find the wavelength of the sound wave.
Ans:
Given that, frequency of sound wave n = 200 Hz
Speed of sound wave v = 400 ms-1
Then the wavelength of sound wave is given by, λ = v/n
Thus, λ = 400/200 = 2m
Hence, the wavelength of sound wave will be 2m.
2) The thunder of cloud is heard 9.8 seconds later than the flash of lightning. If the speed of sound in air is 330 ms–1, what will be the height of the cloud?
Ans:
Given that, the speed of sound in air medium v = 330ms-1
The thunder of cloud is heard 9.8 seconds later than flash of lightning, t = 9.8 seconds
Then, the height of the cloud will be given by
H = v* t = 330* 9.8 = 3234 m
Thus, the height of the cloud will be 3234m.
3) A person who is sitting at a distance of 400 m from a source of sound is listening to a sound of 600 Hz. Find the time period between successive compressions from the source?
Ans:
Given that, frequency of sound n = 600 Hz
We know that, time period = 1/ frequency = 1/ n
T = 1/ 600 = 0.00167 seconds
Thus, the time period between the successive compressions from the source will be 0.00167 seconds.
4) An ultrasonic wave is sent from a ship towards the bottom of the sea. It is found that the time interval between the transmission and reception of the wave is 1.6 seconds. What is the depth of the sea, if the velocity of sound in the seawater is 1400 ms–1?
Ans:
Given that, velocity of sound in sea water v = 1400 ms-1
Time between transmission and reception of the wave t = 1.6 seconds
Then, the speed of sound is given by,
Speed of sound = distance travelled/ time taken
Hence, v = 2s / t
S = v*t/ 2 = 1400*1.6/ 2 = 1120 m
Thus, the depth of the sea will be 1120 m
5) A man is standing between two vertical walls 680 m apart. He claps his hands and hears two distinct echoes after 0.9 seconds and 1.1 second respectively. What is the speed of sound in the air?
Ans:
Given that, distance between two vertical walls d = 680 m
First echo heard after time t1 = 0.9 seconds
Second echo heard after time t2 = 1.1 seconds
Then, the speed of sound in air is given by,
Velocity = distance travelled/ time
Hence, v = 2d/ t
Hence we can write, d = vt/ 2
Now, d1 = vt1/2 and d2= vt2/2
By combining above two equations we get,
Thus, d = v/2* (t1 + t2)
Hence, v= 2d/ (t1 + t2)
V = 2*680/ (0.9 + 1.1)
Thus, v = 2*680 / 2
Hence, v = 680 ms-1
Thus the speed of sound in air is 680 ms-1
6) Two observers are stationed in two boats 4.5 km apart. A sound signal sent by one, under water, reaches the other after 3 seconds. What is the speed of sound in the water?
Ans:
Given that, the distance between two boats is d= 4.5 km = 4500 m
Signal reaches after 3 seconds, t = 3 seconds
Hence, speed of sound in water is,
Speed of sound in water = d/t = 4500/ 3 = 1500 m/s
Hence, the speed of sound in water will be 1500 m/s.
7) A strong sound signal is sent from a ship towards the bottom of the sea. It is received back after 1s.What is the depth of sea given that the speed of sound in water 1450 ms–1?
Ans:
Given that, speed of sound in water v = 1450 m/s
Time required to receive signal back after 1 second, t = 1 second
Then velocity of sound in water = distance travelled / time
Here distance traveled d = depth of the sea.
Hence, depth of sea, d = v*t/ 2 = 1450*1/2 = 725m
Hence, the depth of the sea will be 725 m.
IX.) Answer in Detail
1) What are the factors that affect the speed of sound in gases?
Ans:
In case of gases, following factors affect the velocity of sound waves:
1.) Effect of density:
The velocity of sound waves is inversely proportional to the square root of density of the gas. Hence, as the density of gas is increased the velocity of sound decreases.
2.) Effect of temperature:
The velocity of sound waves is directly proportional to the square root of temperature of the gas. Hence, as the temperature of the gas is increased the velocity of sound waves will also increases.
3.) Effect of relative humidity:
As the humidity of air increases the velocity of sound waves also increases. Hence, in rainy season we can hear a sound clearly from long distance also than on dry day.
2) What is mean by reflection of sound? Explain:
Ans:
Reflection of sound waves:
When sound waves travels in a medium and get bounced back in the same medium by striking to the surface of other medium. This phenomenon is called as reflection of sound.
a) Reflection at the boundary of a rarer medium:
- When the wave traveling through the solid medium strikes the interface between the solid and air then compression will be produced which exerts the force F on the surface of rarer medium.
Due to the less resistance of rarer medium the surface of separation is pushed backwards. Due to the free particles of the rarer medium rarefaction is produced at the interface. In this way, compression is reflected as rarefaction which travels from right to left.
Fig. reflection at the boundary of a denser medium
b) Reflection at the boundary of a denser medium:
- Sound waves travels in a medium in the form of compressions and rarefactions. When the compression traveling from left to right strikes to the rigid wall, it exerts a force F on the rigid wall. Due to which the rigid wall also exerts the opposite reaction on the compression and as a result compression is reflected back.
- In this way, compression is reflected back as a compression in case of reflection from a boundary of denser medium.
Fig. reflection at the boundary of a denser medium
c) Reflection at curved surfaces:
- When sound waves are reflected from plane surfaces it get reflected in the direction as given by laws of reflection of sound. In this type of reflection, the intensity of sound will not be decreased or increased but it remains the same.
- When sound waves are reflected from curved surfaces then the intensity of sound waves changes.
- In case of reflection from convex lens the sound waves are diverged out and due to which intensity of sound waves decreases. While in case of reflection from concave lens, the sound waves are converged and focused at a point at which it is concentrated.
- In case of reflection from parabolic surfaces, the reflected waves are concentrated at a point that’s why many halls are designed with parabolic surfaces.
- And in elliptical surfaces the sound waves are reflected from one focus to another focus.
3) a) What do you understand by the term ‘ultrasonic vibration’?
The ultrasonic waves are the sound waves whose frequency is greater than the 20KHz. Human cannot hear the sound of ultrasonic waves. The creatures such as mosquito, dogs, bat and dolphins can hear detect the ultrasonic waves.
b) State three uses of ultrasonic vibrations.
- Ultrasonic waves are used in cutting and matching of hard materials.
- There are ultrasonic soldering and welding.
- Ultrasonic waves are used in ultrasonography also.
- Ultrasonic waves are also helps in detecting the signals which comes from the developing embryo in the mother’s uterus.
- They are also used to forecast tsunami and earthquakes.
c) Name three animals which can hear ultrasonic vibrations.
The creatures such as mosquito, dogs, bat, and dolphins can detect the ultrasonic waves.
4) a) What is an echo?
Ans:
Echo:
The reproduction of the sound due to the reflection of original sound from different rigid surfaces like walls, ceilings, surfaces of mountains etc. this phenomenon is called as an echo.
b) State two conditions necessary for hearing an echo:
i.) The persistence of hearing for human ears is 0.1 seconds. That means we can hear the sound two times clearly if the time interval between two sounds is at least 0.1 seconds.
The minimum time gap between original sound and an echo must be 0.1 seconds.
ii.) that is to satisfy the first condition, the distance between source of sound and reflecting surface has to satisfy following condition,
Velocity = distance traveled by sound/ time taken
V = 2d/t
And d= vt/2
As t= 0.1 seconds
Then, d= v*0.1/2 = v/20
Thus, the minimum distance required to hear an echo is 1/20th part of the magnitude of sound waves in air.
What are the medical applications of echo?
The principle of echo is used in obstetric ultrasonography, which is used to create real-time visual images of the developing embryo or fetus in the mother’s uterus.
And it is the safe testing tool as it does not uses any harmful radiations.
c) How can you calculate the speed of sound using echo?
To calculate the speed of sound using echo the apparatus required are source of sound pulses, a measuring tape, a sound receiver and a stop watch.
Procedure:
- First we have to measure the distance between source of sound pulse and the reflecting surface with the help of measuring tape as‘d’.
- And a receiver is placed near to the source and sound pulse is emitted from the source.
- By using the stopwatch we have to note the time interval between the instant at which sound is sent and the instant at which echo is received by the receiver. And note it as‘t’.
- By repeating the same procedure for 4-5 times, the average of the time is calculated.
- The total distance travelled by the sound pulse is 2d. Because reaches to the receiver and again reflected back to the source.
- And hence, speed of sound is given by,
Speed of sound = distance travelled/ time taken = 2d/ t
X.) HOT Questions
1.) Suppose that a sound wave and a light wave have the same frequency, then which one has a longer wavelength?
a) Sound b) Light c) both a and b d) data not sufficient
Ans:
a) Light
Because, the wavelength of light wave is longer as its speed is greater than the speed of sound.
2.) When sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source of sound remain the same. Do you hear an echo sound on a hotter day? Justify your answer.
Ans:
- Since, the velocity of sound is directly proportional to the square root of temperature. As temperature increases the velocity of sound also increases.
- On hotter day, the temperature of air is more due to which velocity of sound also increases. As the velocity of sound increases, the time interval between emitted sound and reflected sound will be less.
- But, we know that the minimum time gap between original sound and an echo must be 0.1 seconds.
- Hence, if the time gap between original sound and echo sound will be greater than 0.1 second then echo will be heard.
- And if the time gap between the original sound and echo sound will be less than 0.1 seconds due to increases in speed of sound on hotter day then the echo cannot be heard clearly.