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Echo & Reverberation — Homi Bhabha Exam Physics MCQs with Solutions

Free Homi Bhabha Exam Physics Echo & Reverberation MCQs with step-by-step solutions (6 questions). Part of Sound. Practise online on Prepizo — no login needed.

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Questions with solutions

Q1 — Echo & Reverberation · medium · theory
An echo is heard because sound is:
A. transmitted through the obstacle
B. absorbed by the obstacle
C. reflected back from a distant obstacle  ✓ Correct
D. refracted through the air
Solution: An echo is the repetition of a sound caused by its reflection from a surface.
Q2 — Echo & Reverberation · hard · numerical
A person shouts towards a cliff and hears the echo after 3 s. If the speed of sound is 340 m/s, the distance of the cliff is:
A. 113 m
B. 1020 m
C. 340 m
D. 510 m  ✓ Correct
Solution: Sound travels to the cliff and back: total distance $= 340 \times 3 = 1020$ m, so the cliff is $\dfrac{1020}{2} = 510$ m away.
Q3 — Echo & Reverberation · hard · theory
To hear a distinct echo, the reflecting surface must be at least about 17 m away because:
A. echoes need bright light
B. the surface must be very hard
C. sound travels very slowly
D. the ear can distinguish two sounds separated by at least 0.1 s  ✓ Correct
Solution: The ear resolves sounds ~0.1 s apart; in 0.1 s sound travels 34 m (there and back), so the surface must be at least ~17 m away.
Q4 — Echo & Reverberation · medium · theory
The persistence of sound in a large hall due to repeated reflections is called:
A. diffraction
B. refraction
C. an echo
D. reverberation  ✓ Correct
Solution: Reverberation is the prolonging of sound by multiple reflections before it dies out.
Q5 — Echo & Reverberation · medium · theory
The ceilings and walls of cinema halls are made of sound-absorbing materials to:
A. reduce reverberation  ✓ Correct
B. reflect all sound
C. increase reverberation
D. increase the echo
Solution: Absorbing materials cut down repeated reflections, reducing reverberation so speech is clear.
Q6 — Echo & Reverberation · hard · numerical
A ship sends an ultrasound pulse to the sea bed and receives the echo after 2 s. If the speed of sound in water is 1500 m/s, the depth of the sea is:
A. 1500 m  ✓ Correct
B. 1000 m
C. 750 m
D. 3000 m
Solution: Total path $= 1500 \times 2 = 3000$ m; depth $= \dfrac{3000}{2} = 1500$ m (this method is SONAR).