NEET Foundation Physics Chapter 5 Sound Short Answer Questions

Chapter 5 Sound Short Answer Type Question And Answers

Question 1. If an echo is heard 5 seconds later in a theatre, calculate the distance of the reflecting surface.
Answer.

Given:

If an echo is heard 5 seconds later in a theatre

Total distance travelled by sound wave,

D = 5 × 340 = 1700 m

Distance of the reflecting surface

= D/2 = 1700/2 = 850 m

Distance of the reflecting surface = 850 m

Question 2. Distinguish between intensity and loudness.
Answer.

Difference between intensity and loudness:

Question 3. What are the requirements of the medium for the sound to propagate?
Answer.

Requirements of the medium for the sound to propagate:

The medium required for propagation of sound must have the following properties:

The medium must be elastic so that its particles may come back to their initial positions after displacement on either side.

The medium must have inertia so that its particles may store mechanical energy.

The medium should be frictionless so that there is no loss of energy in propagation of sound through it.

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Question 4. A tuning fork having a frequency of 250 Hz is made to vibrate by striking it against a rubber pod. What is the wavelength of the sound waves it produces in

1. air  
2. water
3. iron?

Velocity of sound in air, water and iron = 330 ms^-1, 1500 ms^-1 and 5 kms^-1 respectively.
Answer.

Given

A tuning fork having a frequency of 250 Hz is made to vibrate by striking it against a rubber pod.

Velocity of sound in air, water and iron = 330 ms^-1, 1500 ms^-1 and 5 kms^-1 respectively.

We know that

v = fλ

i.e., λ = \(\frac{v}{f}\)

In air, \(\lambda_a=\frac{330 \mathrm{~ms}^{-1}}{250 \mathrm{~s}^{-1}}\)

= 1.32m

Similarly,

\(\lambda_w=\frac{1500 \mathrm{~ms}^{-1}}{250 \mathrm{~s}^{-1}}\)

= 6.0m

In iron,

\(\lambda_{\text {iron }}=\frac{5000 \mathrm{~ms}^{-1}}{250 \mathrm{~s}^{-1}}\)

= 20.0 m

Question 5. If a tuning fork having a frequency of 400 Hz is made to vibrate continuously how many

1. Compression
2. Rarefactions

Pass a point P in air in 0.1 s?
Answer.

Given

A tuning fork having a frequency of 400 Hz is made to vibrate continuously

When a tuning fork vibrates once, it produces one sound wave, which consists of one compression and one rarefaction.

∴ Its frequency is 400 Hz, it vibrates 400 times in one second.

In one second, it vibrates 400 times.

So, in 0.1 seconds it vibrates 0.1 × 400 = 40 times.

So, it will produce 40 waves, i.e., 40 compressions and 40 rarefactions.

∴ Number of compressions passing a point P  in 0.1 second is 40.

Similarly, a number of rarefaction passing a point P is 0.1 second is also 40.

Question 6. Displacement–The distance graph of particles vibrating in a gas under the influence of a tuning fork of frequency 512 Hz is as follows: Find the velocity of sound in the gas.

Answer.

Given

Points A and B represent the consecutive compressions. Distance between them is called ‘wavelength’ (λ).

So, λ = (100 cm − 20 cm) = 80 cm

Distance between points C and D = (120 cm − 40 cm) = 80 cm is also the wavelength(λ).

∴ λ = 80 cm = 0.8 m

We know that

v = fλ

= 512 × 0.8

= 409.6 ms^-1

∴ Velocity of sound in the gas = 409.6 ms^-1

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Question 7. A train is travelling in a railway track made of iron. A boy standing far away puts his ears to the track. He hears two sounds. Why? Calculate the time interval between these two sound waves.
Velocity of sound in air = v_a
Velocity of sound in iron = v_s (where v_s > v_a)
Answer.

Given

A train is travelling in a railway track made of iron. A boy standing far away puts his ears to the track. He hears two sounds.

It is because he receives the sound waves travelling through air as well as through iron. Since sound travels much faster through iron he will the hear sound travelling through iron first.

Time ‘t1’ taken by the sound to travel from the train to the boy through the iron rail is given by

\(\text { Time }=\frac{\text { Distance }}{\text { Speed }}\)

\(t_1=\frac{d}{v_n}\)  (1)

Similarly, time ‘t2’ taken by sound to travel to him is,

\(t_2=\frac{d}{D_d}\)  (2)

∴ Time interval ∆t between two sound waves reaching the boy is

∆t = t₂ − t₁

= \(\left(\frac{d}{v_a}-\frac{d}{v_n}\right)\)

Question 8. Calculate the velocity of sound through iron given that the modulus of elasticity for iron = 2 × 10¹¹ Pa and density of iron = 8 × 10³ kgm^-3.
Answer.

Velocity of sound through a solid is given by

\(v=\sqrt{\frac{Y}{\rho}}\)

Where Y = young’s modulus

ρ = Density

V = \(\sqrt{\frac{\left(2 \times 10^{11}\right) \mathrm{Pa}}{\left(8 \times 10^3\right) \mathrm{kgm}^{-3}}}\)

= \(\sqrt{0.25 \times 10^8 \frac{\mathrm{Nm}^{-2}}{\mathrm{kgm}^{-3}}}\)

= \(\sqrt{25 \times 10^6 \frac{\mathrm{kgms}^{-2} \mathrm{~m}^{-2}}{\mathrm{kgm}^{-3}}}\)

= \(\sqrt{25 \times 10^6 \mathrm{~m}^2 \mathrm{~s}^{-2}}\)

= 5 x 10³ ms^-1

The velocity of sound = 5 x 10³ ms^-1

Question 9. How does a SONAR work?
Answer:

Working Of SONAR

SONAR is a device which is used in the ships to locate rocks, icebergs submarines, old sank in the seas etc. Ultrasonic sound of high frequency are sent from a ship on the surface. The waves travel in the straight line till they hit something like shipwreck or submarine.

On hitting the body these waves are reflected. The transmitter sending the waves note the time lag between sending the signal and receiving it back. This time lag is multiplied by speed of sound in water and the distance calculated is halved to get the actual distance of the object from the ship.

Question 10. Describe the structure and function of a human ear.
Answer:

Structure and function of a human ear:

The human ear consists of three parts:

1. Outer ear: It is functionally the simplest part of the ear. It consists of ‘pinna’ or auricle i.e. the visible portion of the ear, the external acoustic meatus i.e. the outside opening to the ear canal and the ear canal.
2. Middle ear: The middle ear consists of eardrum or tympanic membrane connected at the end of auditory canal. The eardrum vibrates when compression and rarefaction of the sound waves hit it. The three bones (hammer, anvil and stirrup) present in the middle ear the pressure variations several times. Thus the middle ear transmits the sound wave’s amplified pressure variation to the inner ear.
3. Inner ear: The inner ear converts the sound waves amplified pressure variation into electrical signals. This work is done by cochlea, a snail-shaped organ. The cochlea is filled with water like fluid and its inner surface has large number of hair like nerve cells. The amplified pressure variation produces vibrations in the nerve cells and they in turn release electrical signals which are sent to the brain along the auditory nerve. The brain interprets the electrical signals through a complex process as sounds.

Question 11. What are the factors affecting the speed of sound in a gas?
Answer:

Speed of sound in a gas is affected by:

1. Density
2. Temperature
3. Humidity
4. Direction of wind

Question 12. What are the factors which does not affect the speed of sound in air?
Answer:

The factors which does not affect the speed of sound in air are:

1. Wavelength
2. Frequency
3. Amplitude
4. Pressure

Question 13. What are the laws of reflection of sound?
Answer:

The laws of reflection of sound are:

• The angle of incidence is equal to angle of reflection.
• The incident ray, reflected ray and normal at the point of incidence, all lie in the same plane.

Question 14. What do you mean by reverberation?
Answer:

Reverberation:

When echo is heard multiple times, due to repeated and multiple reflections of sound from different reflecting surfaces, it causes persistence of sound. This phenomenon is called reverberation.

Question 15. Name the devices which use multiple reflections of sound.
Answer:

Loudspeaker, megaphone, soundboard and stethoscope are the devices which use multiple reflections of sound.

Question 16. A boy shouted at the top of the mountain. The echo is heard after 10 seconds. Speed of the sound is 450 m/s. How far is the mountain from the boy?
Answer:

Given

A boy shouted at the top of the mountain. The echo is heard after 10 seconds. Speed of the sound is 450 m/s.

v = 450 m/s

T = 10 s

Distance d = 450 × 10 = 4500 m

As sound has to travel twice the distance, hence distance between boy and mountain = 4500/2

= 2250 m

Hence distance between boy and mountain = 2250 m

Question 17. Why sound is called a mechanical wave?
Answer:

Mechanical wave:

Sound is called a mechanical wave because it requires a medium to travel from one place to another.

Question 18. What properties should the medium have for the propagation of sound?
Answer:

The medium required for propagation of sound must have the following properties:

• The medium should be elastic so that its particles comes back to the mean position after displacement on the either side.
• The medium must have inertia so that its particles may store mechanical energy.
• The medium should be frictionless so that there is no loss of energy in propagation of sound through it.

Question 19. What do you mean by crest and trough?
Answer:

Crest And Trough:

The position of maximum upward displacement is called crest and position of maximum downward displacement is called trough. These two are created in the medium during the propagation of transverse wave.

Question 20. What is the difference between progressive wave and stationary wave?
Answer:

The difference between progressive wave and stationary wave

Progressive waves start at a point and moves indefinitely and infinitely to all parts of the medium. These waves transmit energy from one place to another. While standing waves appear to stand at a place and confined between two points in a medium. These waves store energy in them.

Question 21. What are the characteristics of a sound wave?
Answer:

The characteristics of the sound waves are:

• Amplitude
• Time period
• Frequency
• Wavelength
• Wave velocity

Question 22. On which factors does the speed of a sound in a medium depends?
Answer:

Speed of sound in a medium depends on elasticity of the medium as well as density of the medium.

Question 23. A sound wave has a frequency of 200 Hz. Wavelength is 15 m. How much time it will take to cover 2 km?
Answer:

Given

A sound wave has a frequency of 200 Hz. Wavelength is 15 m.

ν = 200 Hz

λ = 15 m

Speed v = ν × λ = 200 × 15 = 3000 m/s

Time t = d/υ

= 2000/3000

= 0.67 seconds

0.67 seconds time will it take to cover 2 km

Question 24. A sound has a frequency of 220 Hz. Its speed is 440 m/s. What will be the wavelength of the sound wave?
Answer:

Given

A sound has a frequency of 220 Hz. Its speed is 440 m/s.

ν = 220 Hz

v = 440 m/s

λ = v/ν = 440/220 = 2 m

The wavelength of the sound wave λ = 2 m