Chapter 7ICSE Class 10 Physics100% Free

Sound — ICSE Class 10 Physics Important Questions

13 ICSE Class 10 Physics practice questions on Sound, each with a full model answer, covering 6 topics from the chapter in the question formats used in the exam.

By The Classmate AI Editorial Team

Reviewed by Classmate AI Team · 17 October 2026

Last updated 25 October 2026

13
Questions
6
Topics
3
Key concepts
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With answers
Quick answer

ICSE Sound questions cover reflection of sound and echoes (using v=2dtv=\dfrac{2d}{t}v=2d/t, the minimum 17 m17\,\text{m}17 m condition and SONAR), natural, damped, forced and resonant vibrations with everyday examples, and the characteristics of sound: loudness and intensity, pitch (frequency), quality (waveform), and sound level in decibels with noise pollution. Expect echo numericals and reason-based questions on resonance.

About Sound

In this ICSE Class 10 Physics chapter Sound you learn how sound waves are reflected to produce echoes and the conditions and calculations involved, the difference between natural, forced and resonant (sympathetic) vibrations, and the phenomenon of resonance with its examples and demonstrations. You also learn the three characteristics of a musical sound - loudness, pitch and quality (timbre) - and how each depends on amplitude, frequency and waveform respectively.

Reflection of sound, echoes and SONARNumericals on speed of sound and echoesNatural, damped, forced and resonant vibrationsResonance and its demonstrationsLoudness, intensity, pitch and qualitySound level in decibels; noise pollution

Key concepts & formulas

Echo and reflection of sound

An echo is a distinct repetition of sound heard after reflection from a distant surface. To hear a separate echo the reflector must be at least 17 m17\,\text{m}17 m away, because the persistence of hearing is 0.1 s0.1\,\text{s}0.1 s: d=v×t2=340×0.12=17 md=\dfrac{v\times t}{2}=\dfrac{340\times0.1}{2}=17\,\text{m}d=(v× t)/2=340×0.1/2=17 m.

Types of vibration

Natural (free) vibration occurs at a body's own natural frequency once disturbed. Forced vibration is set up by an external periodic force at the force's frequency. Resonance is a special forced vibration where the applied frequency equals the body's natural frequency, giving a large amplitude.

Characteristics of a musical sound

Loudness depends on amplitude (and area of the vibrating body); pitch depends on frequency (higher frequency gives a shriller note); quality (timbre) depends on the waveform, letting us distinguish two instruments sounding the same note.

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Important questions with answers

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Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

The characteristic of sound that distinguishes a shrill note from a flat note is its:

  1. (a)

    loudness

  2. (b)

    pitch

  3. (c)

    quality

  4. (d)

    speed

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Answer: (b) pitch.

Pitch depends on frequency; a higher frequency gives a shriller (higher-pitched) note and a lower frequency a flatter note.

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Q2MCQEasy1 mark

The sound level of a noise is expressed in:

  1. (a)

    hertz (Hz)

  2. (b)

    decibel (dB)

  3. (c)

    metre per second

  4. (d)

    watt

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Answer: (b) decibel (dB). Ordinary conversation is about 60 dB60\,\text{dB}60 dB. Long exposure to sound above about 80 dB80\,\text{dB}80 dB, such as heavy traffic or loudspeakers, is noise pollution: it can cause headaches, stress and loss of hearing. The hertz is the unit of frequency.

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Q3MCQModerate1 mark

When a vibrating tuning fork is placed on a table, the sound becomes louder. This is an example of:

  1. (a)

    natural vibration

  2. (b)

    forced vibration

  3. (c)

    resonance

  4. (d)

    an echo

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Answer: (b) forced vibration.

The fork forces the table top to vibrate at the fork's frequency. The larger vibrating area produces a louder sound, but since the table's natural frequency need not equal the fork's, it is forced (not resonant) vibration.

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Q4MCQHOTS1 mark

A person stands between two parallel cliffs and fires a gun. He hears the first echo after 2 s2\,\text{s}2 s and the next after 3 s3\,\text{s}3 s. If the speed of sound is 340 m s−1340\,\text{m s}^{-1}340 m s^-1, the distance between the two cliffs is:

  1. (a)

    340 m340\,\text{m}340 m

  2. (b)

    510 m510\,\text{m}510 m

  3. (c)

    850 m850\,\text{m}850 m

  4. (d)

    1020 m1020\,\text{m}1020 m

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Answer: (c) 850 m850\,\text{m}850 m.

First cliff: d1=340×22=340 md_1=\dfrac{340\times2}{2}=340\,\text{m}d₁=340×2/2=340 m. Second cliff: d2=340×32=510 md_2=\dfrac{340\times3}{2}=510\,\text{m}d₂=340×3/2=510 m. Distance between cliffs =d1+d2=340+510=850 m=d_1+d_2=340+510=850\,\text{m}=d₁+d₂=340+510=850 m.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): Soldiers are ordered to break their steps while marching across a suspension bridge.

Reason (R): If the frequency of marching happens to equal the natural frequency of the bridge, resonance can set the bridge into large-amplitude vibrations and it may collapse.

  1. (a)

    Both A and R are true and R is the correct explanation of A

  2. (b)

    Both A and R are true but R is not the correct explanation of A

  3. (c)

    A is true but R is false

  4. (d)

    A is false but R is true

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Answer: (a) Rhythmic marching could match the bridge's natural frequency and cause resonance, building up dangerously large vibrations. Breaking step prevents this, so R correctly explains A.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

Distinguish between natural vibrations and damped vibrations, giving one example of each.

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Natural (free) vibrations: a body vibrates at its own natural frequency with a constant amplitude, with no resistance acting. Example: a simple pendulum swinging in vacuum (or a struck tuning fork, very nearly).

Damped vibrations: the amplitude decreases gradually with time, because friction or air resistance takes away energy; the vibrations finally stop. Example: a pendulum swinging in air slowly comes to rest; the sound of a struck tuning fork dies away.

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Q7Very ShortModerate2 marks

State two differences between forced vibrations and resonance.

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Forced vibration: the body vibrates under an external periodic force at the frequency of the applied force, which may differ from the body's natural frequency; the amplitude is generally small.

Resonance: a special case of forced vibration in which the frequency of the applied force equals the natural frequency of the body; the amplitude of vibration becomes very large. Thus every resonance is a forced vibration, but not every forced vibration is resonance.

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Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

A man standing 85 m85\,\text{m}85 m from a tall wall claps his hands and hears the echo 0.5 s0.5\,\text{s}0.5 s later. Calculate the speed of sound in air.

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The sound travels to the wall and back, so it covers twice the distance to the wall.

Total distance =2d=2×85=170 m=2d=2\times85=170\,\text{m}=2d=2×85=170 m.

Time t=0.5 st=0.5\,\text{s}t=0.5 s.

v=total distancetime=2dt=1700.5=340 m s−1v=\dfrac{\text{total distance}}{\text{time}}=\dfrac{2d}{t}=\dfrac{170}{0.5}=340\,\text{m s}^{-1}v=(total distance)/time=2d/t=170/0.5=340 m s^-1

The speed of sound in air is 340 m s−1340\,\text{m s}^{-1}340 m s^-1.

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Q9Short AnswerModerate3 marks

Define (i) loudness, (ii) pitch and (iii) quality of a sound, and state the physical quantity on which each depends.

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(i) Loudness: the characteristic by which a loud sound is distinguished from a faint one. It depends on the amplitude of vibration (and on the area of the vibrating body and distance from the source). Loudness is subjective (it depends on the listener's ear), whereas intensity, the sound energy passing per second through unit area, is objective and can be measured.

(ii) Pitch: the characteristic by which a shrill (high) note is distinguished from a flat (low) note. It depends on the frequency of the sound - higher frequency means higher pitch.

(iii) Quality (timbre): the characteristic that enables us to distinguish between two sounds of the same loudness and pitch produced by different sources. It depends on the waveform (the number and relative intensity of overtones).

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Q10Short AnswerHOTS3 marks

A ship sends a sound signal towards the sea bed and receives its echo after 3 s3\,\text{s}3 s. If the speed of sound in sea water is 1500 m s−11500\,\text{m s}^{-1}1500 m s^-1, find the depth of the sea. Why is sound, and not light, used for this purpose?

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The sound travels down to the sea bed and back, so it covers twice the depth.

2d=v×t=1500×3=4500 m2d=v\times t=1500\times3=4500\,\text{m}2d=v× t=1500×3=4500 m

d=45002=2250 md=\dfrac{4500}{2}=2250\,\text{m}d=4500/2=2250 m

The depth of the sea is 2250 m2250\,\text{m}2250 m.

Sound is used (in SONAR) rather than light because sound waves can travel long distances through water with little absorption and are strongly reflected by the sea bed, whereas light is quickly absorbed and scattered by water and cannot penetrate to such depths.

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Long answer questions (5 marks)

Q11Long AnswerModerate5 marks

(a) What is resonance? (b) Describe an experiment with two identical pendulums (or tuning forks on sound boxes) to demonstrate resonance. (c) Give one useful and one harmful example of resonance in daily life.

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(a) Resonance is the phenomenon in which a body is set into vibrations with a large amplitude when it is acted upon by a periodic force whose frequency is equal to the natural frequency of the body.

(b) Take two identical tuning forks AAA and BBB of the same frequency, each mounted on its own sound box, and place the open ends of the boxes facing each other. Sound fork AAA with a rubber hammer and then quickly damp it with the hand. Fork BBB is now found to be vibrating and giving out sound. Because both forks have the same natural frequency, the periodic air pressure variations from AAA force BBB into resonant vibration. (Two identical pendulums coupled on a common string behave similarly - energy transfers to the one having the same length/natural frequency.)

(c) Useful: resonance in the air column of musical instruments (flute, sitar) and in a radio-tuning circuit that selects a station. Harmful: resonance can cause a bridge to collapse under rhythmic marching or high winds, and can shatter a glass when a singer's note matches its natural frequency (any one of each).

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Q12Long AnswerHOTS5 marks

(a) State the conditions necessary to hear a distinct echo. (b) A boy standing in front of a cliff fires a gun and hears the echo after 2.5 s2.5\,\text{s}2.5 s. He then moves 170 m170\,\text{m}170 m towards the cliff and fires again, hearing the echo after 1.5 s1.5\,\text{s}1.5 s. Calculate the speed of sound and the original distance of the boy from the cliff.

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(a) Conditions for a distinct echo: (i) the minimum distance between the source and the reflecting surface should be 17 m17\,\text{m}17 m (so the reflected sound returns after at least 0.1 s0.1\,\text{s}0.1 s, the limit of persistence of hearing); (ii) the reflecting surface should be large and hard; and the intensity of sound should be sufficient to be heard after reflection.

(b) Let the original distance be ddd and the speed of sound be vvv.

First firing: 2d=v×2.52d=v\times2.52d=v×2.5 ... (1)

After moving 170 m170\,\text{m}170 m closer, distance =(d−170)=(d-170)=(d-170): 2(d−170)=v×1.52(d-170)=v\times1.52(d-170)=v×1.5 ... (2)

Subtract (2) from (1): 2d−2(d−170)=v(2.5−1.5)2d-2(d-170)=v(2.5-1.5)2d-2(d-170)=v(2.5-1.5)

340=v(1.0)⇒v=340 m s−1340=v(1.0)\Rightarrow v=340\,\text{m s}^{-1}340=v(1.0)⇒ v=340 m s^-1.

From (1): 2d=340×2.5=8502d=340\times2.5=8502d=340×2.5=850, so d=425 md=425\,\text{m}d=425 m.

The speed of sound is 340 m s−1340\,\text{m s}^{-1}340 m s^-1 and the boy was originally 425 m425\,\text{m}425 m from the cliff.

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Case-based questions (4 marks)

Q13Case-basedModerate4 marks

A science teacher demonstrates that a vibrating tuning fork of a certain frequency held over an open pipe partly filled with water produces a very loud sound only when the water is adjusted to a particular level.

(i) Name the phenomenon responsible for the loud sound.

(ii) State the condition satisfied by the air column at that particular level.

(iii) The teacher then strikes the same fork harder. Which characteristic of the sound changes, and why?

(iv) If a tuning fork of higher frequency were used, how would the pitch of the sound compare?

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(i) The loud sound is produced by resonance (the air column resonates with the tuning fork).

(ii) At that level the natural frequency of the vibrating air column becomes equal to the frequency of the tuning fork, so the air column vibrates with maximum (large) amplitude and the sound is loudest.

(iii) Striking the fork harder increases the amplitude of vibration, so the loudness of the sound increases. The frequency (pitch) stays the same because it is fixed by the fork.

(iv) A tuning fork of higher frequency produces a note of higher pitch (a shriller sound), since pitch increases with frequency.

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All ICSE Class 10 Physics Chapters

Frequently asked questions

  • What are the important questions in Sound for ICSE Class 10?
    Echo numericals and the conditions to hear an echo, SONAR depth problems, natural, damped, forced and resonant vibrations, and loudness, pitch and quality. This page has 13 questions with answers, including a resonance experiment and a tuning-fork-over-a-pipe case study.
  • Why must a reflecting surface be at least 17 m away to hear an echo?
    The ear keeps a sound for about 0.10.10.1 s, so the echo must arrive at least 0.10.10.1 s after the original. Sound travels to the wall and back, so d=vt2=340×0.12=17d=\dfrac{v t}{2}=\dfrac{340\times0.1}{2}=17d=(v t)/2=340×0.1/2=17 m in air.
  • What is the difference between forced vibration and resonance?
    In forced vibration a body vibrates at the frequency of an external periodic force, usually with small amplitude. Resonance is the special case where that frequency equals the body's natural frequency, so the amplitude becomes large — why soldiers break step on a suspension bridge.
  • How do you calculate the speed of sound from an echo?
    Use v=2dtv=\dfrac{2d}{t}v=2d/t, since the sound covers the distance twice. A clap 858585 m from a wall with the echo heard after 0.50.50.5 s gives v=2×850.5=340 m s−1v=\dfrac{2\times85}{0.5}=340\ \text{m s}^{-1}v=2×85/0.5=340 m s^-1. SONAR depth uses the same idea: d=vt2d=\dfrac{vt}{2}d=vt/2.
  • Is there a Class 9 ICSE chapter on sound to revise first?
    Yes — Propagation of Sound Waves (ICSE Class 9 Physics, Chapter 8) covers wave basics and the speed of sound. In Class 10, Sound is Chapter 7, after Spectrum. Each of these two chapters has its own important-questions page here.
  • What should I focus on to score well in ICSE Sound?
    Learn which property each characteristic depends on — loudness on amplitude, pitch on frequency, quality on waveform — and practise echo numericals without forgetting the factor of 222. Reason questions on resonance carry marks for naming the exact condition, equal frequencies.
  • Where can I get Sound Class 10 ICSE important questions as a PDF?
    On this page: enter your email in the free PDF box and a printable worksheet of all 13 Sound questions with answers opens, ready to save as a PDF.
  • Can I take an online test on ICSE Sound?
    Yes. After a free sign-up, Create Test lets you pick ICSE Physics and Sound and set the difficulty and timer; results go to your SWOT report. The Free plan includes two practice sets a week, and the 7-day Premium trial removes the limit.

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