Sound MCQ Type Questions for ICSE Physics Class-10

Sound MCQ Type Questions for ICSE Physics Class-10 . These MCQ  / Objective Type Questions is based on latest reduced syllabus according 2021-22 session on bifurcated pattern. Main motto of MCQ Type Question is cracking the next upcoming exam of council. Visit official website CISCE for detail information about ICSE Board Class-10 Physics.

Sound for ICSE Physics Class-10 MCQ Type Questions

Board ICSE
Class 10th (X)
Subject Physics
Chapter Sound
Syllabus  on bifurcated syllabus (after reduction)
Session 2021-22
Topic MCQ / Objective Type Question

MCQ Type Questions of Sound for ICSE Class-10 Physics

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Question 1: The minimum distance between the source and the reflector in air, so that an echo is heard is approximately equal to:

(a) 10 m

(b) 17 m

(c) 34 m

(d) 50 m

Answer (b) 17 m

Explanation: An echo is heard distinctly if it reaches the ear at least 0.1 s after the original sound.

If d is the distance between the observer and the obstacle and V is the speed of sound, then the total distance travelled by the sound to reach the obstacle and then to come back is 2d and the time taken is,

t = Total distance travelled/Speed of sound = 2d/V

or, d = V t/2

Putting t = 0.1 s and V = 340 m/s in air at ordinary temperature, we get:

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d = (340 x 0.1)/2 = 17 m

Thus, to hear an echo distinctly, the minimum distance between the source and the reflector in air is 17 m.

Question 2: To detect the obstacles in their path, bats produce:

(a) Infrasonic waves

(b) Ultrasonic waves

(c) Electromagnetic waves

(d) Radio waves

Answer (b) Ultrasonic waves

Question 3: The wavelength of waves produced on the surface of water is 20cm. If the wave velocity is 24m s-1, calculate (i) the number of waves produced in one second and (ii) the time in which one wave is produced.

(a) 60 and  8.3 x 10-3 seconds

(b) 120 and  9.5 x 10-3 seconds

(c) 120 and  8.3 x 10-3 seconds

(d) 140 and  8.3 x 10-3 seconds

Answer (c) 120 and  8.3 x 10-3 seconds 

(i)Frequency or the number of waves produced per second

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= Velocity/Wavelength

= 24 / 20 x 10-2

=120

(ii)Time = 1/ frequency = 1/ 120= 8.3 x 10-3 seconds

Question 4: Calculate the minimum distance in air required between the source of sound and the obstacle to hear an echo. Take the speed of sound in air = 350m s-1

(a) 16.5 m

(b) 34.5 m

(c) 17.5 m

(d) none of these

Answer (c) 17.5 m

Velocity = 2D/Time

350 = 2 x D/ 0.1

D =350 x 0.1 / 2 = 17.5 m

Question 5: What should be the minimum distance between the source and reflector in water so that echo is heard distinctly? (The speed of sound in water = 1400m/s)

(a) 70 m

(b) 80 m

(c) 50 m

(d) none of these

Answer (a) 70 m

Velocity = 2D/Time

1400 = 2 x D/ 0.1

D = 1400 x 0.1/ 2 = 70 m

Question 6:  A man standing 25 m away from a wall produces a sound and receives the reflected sound. Calculate the time after which he receives the reflected sound if the speed of sound in air is 350m/s-1

(a) 1.143 seconds

(b) 0.143 seconds

(c) 0.15 seconds

(d) none of these

Answer (b) 0.143 seconds

Velocity = 2D/Time

Time = 2 x 25 / 350 =0.143 seconds

Question 7:  A man standing 48 m away from a wall fires a gun. Calculate the time after which an echo is heard. (The speed of sound in air is 320m/s-1).

(a)  0.3 seconds

(b)  0.5 seconds

(c)  1.3 seconds

(d) none of these

Answer (a)  0.3 seconds

Velocity = 2 x D/Time

Time after which an echo is heard = 2 D/Velocity = 2 x 48 / 320 = 0.3 seconds

Question 8:  A pendulum has a frequency of 5 vibrations per second. An observer starts the pendulum and fires a gun simultaneously. He hears an echo from the cliff after 8 vibrations of the pendulum. If the velocity of sound in air is 340m/s-1 , find the distance between the cliff and the observer.

(a)  372 m

(b)  275 m

(c)  350 m

(d)  272 m

Answer (d)  272 m

5 vibrations by pendulum in 1 sec

So8 vibrations in 8/5 seconds = 1.6 sec

Velocity = 2 x D/ time

340 = 2 x D/ 1.6

D = 340 x 1.6 / 2 = 272 m

Question 9:  A person standing between two vertical cliffs produces the sound. Two successive echoes are heard at 4s and 6s. Calculate the distance between the cliffs. (Speed of sound in air = 320m/s)

(a) 3200 m

(b) 800 m

(c) 1600 m

(d) 1900 m

Answer (c) 1600 m

The distance of first cliff from the person, 2 x D1 = velocity x time

D1 = 320 x 4 / 2 = 640 m

Distance of the second cliff from the person, D2 = 320 x 6 / 2 = 960 m

Distance between cliffs = D1 + D2 = 640 + 960 = 1600 m

Question 10: SONAR is a technique used to. ………………

(a) detect the submarine inside sea
(b) determine the size of blue whale
(c) detect the impurities present in sea water
(d) detect the depth of sea bed

Answer (d) detect the depth of sea bed

Question 11: Bats can catch their prey by using the principle of ……….

(a) reflection of sound
(b) refraction of sound
(c) concept of echo
(d) both (a) and (c)

Answer (d) both (a) and (c)

Question 12: Modern super sonicC war planes use the principle of ……………. while detecting their targets.

(a) echo

(b) SONAR

(c) reverberation

(d) none of these

Answer (a) echo

Question 13: The spread or a glacier in the artic region can be judged by a submarine by using the principle of

(a) repetition of sound
(b) reflection of sound
(c) refraction of sound
(d) none of the above

Answer (b) reflection of sound

Question 14: The repetition of sound heard in a cave or empty room is a case of …………….

(a) echo

(b) superposition

(c) interpolation

(d) reverberation.

Answer (d) reverberation.

Question 15: On sending an ultrasonic wave from a ship towards the bottom of a sea, the time interval between sending the wave and receiving it back is found to be 1.5s. If the velocity of wave in sea water is 1400m/s, find the depth of sea.

(a) 1000 m

(b) 1050 m

(c) 1150 m

(d) none of the these

Answer (b) 1050 m Depth of the sea = velocity x time/2 = 1400 x 1.5 / 2 = 1050 m

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