Class 9 Science Exploration Chapter 10 Solutions (NCERT 2026–27) – Sound Waves: Characteristics and Applications

These Class 9 Science Exploration Chapter 10 solutions cover Sound Waves: Characteristics and Applications from the new NCF-2023 textbook (2026–27).

Class: 9 Subject: Science Book: Exploration Chapter: 10 Exercise: Revise, Reflect, Refine (15 Qs) Session: 2026–27
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Class 9 Science Exploration Chapter 10 Solutions – Overview

Chapter 10 of Exploration, Sound Waves: Characteristics and Applications, explains how sound is produced by vibrating bodies and travels as a longitudinal mechanical wave through a medium as a series of compressions and rarefactions. It covers the characteristics of a sound wave — wavelength, frequency, time period, amplitude and speed — the key relation v = f × λ, the ideas of echo and reverberation, and important applications such as SONAR and ultrasound. These Class 9 Science Exploration Chapter 10 solutions answer every textbook question step by step.

Key Concepts & Definitions

Sound is a longitudinal mechanical wave — it needs a medium and the particles vibrate along the direction of travel.

Compression: a high-density, high-pressure region; rarefaction: a low-density, low-pressure region.

Wavelength (λ): distance between two consecutive compressions (or rarefactions). Frequency (f): number of waves passing a point per second (Hz). Time period (T): time for one wave = 1/f. Amplitude: the maximum change in density/pressure.

Echo: a distinct reflected sound heard when the gap is at least 0.1 s. Reverberation: persistence of sound by repeated reflections when the gap is less than 0.1 s.

SONAR & ultrasound use high-frequency sound to measure distances and form images.

Sound Wave Formulas

v = f × λ (speed = frequency × wavelength).

T = 1/f (time period is the reciprocal of frequency).

Echo distance: 2d = v × t (the sound travels to the obstacle and back).

Echo is heard only if the time gap t ≥ 0.1 s.

“Think It Over” — Answers

Two astronauts standing close together in space cannot hear each other speak directly. Why?

ANSWERSound is a mechanical wave and needs a material medium to travel. Space is almost a vacuum with no air, so the sound cannot propagate — the astronauts must use radios, which work with electromagnetic waves that can travel through vacuum.

How is sound produced, and how does it reach our ears?

ANSWERSound is produced by a vibrating object. The vibrations push and pull the surrounding air, creating compressions and rarefactions that travel out as a longitudinal wave; on reaching our ears they make the eardrum vibrate, and we hear the sound.

Class 9 Science Exploration Chapter 10 Solutions — Revise, Reflect, Refine

1. Which observation best supports the idea that sound is a mechanical wave? (i) Sound shows reflection (ii) Sound needs a medium to propagate (iii) Sound has frequency (iv) Sound carries energy

ANSWERCorrect option: (ii) Sound needs a medium to propagate. A mechanical wave can travel only through a material medium — this is what distinguishes it from electromagnetic waves, which travel through vacuum.

2. For a sound wave in a medium, increasing its frequency will increase its (i) wavelength (ii) speed (iii) number of compressions per second (iv) time period

ANSWERCorrect option: (iii) number of compressions per second. Frequency is the number of compressions passing per second; in the same medium the speed stays the same, so a higher frequency means a shorter wavelength and a smaller time period.

3. If 20 compressions pass a point in 4 seconds, the frequency is (i) 80 Hz (ii) 5 Hz (iii) 10 Hz (iv) 0.2 Hz

ANSWERCorrect option: (ii) 5 Hz. Frequency = number of waves / time = 20 / 4 = 5 Hz.

4. In a room, the reflected sound reaches the ear 0.05 s after its production. Will it produce an echo or reverberation? Justify.

ANSWER An echo can be heard only if the reflected sound reaches the ear at least 0.1 s after the original sound. Here the gap is only 0.05 s, which is less than 0.1 s, so the reflected sound merges with the original — it produces reverberation, not an echo.

5. Graphs of two sound waves are given in Fig. 10.30 (same scales on both axes). Which wave has (i) greater wavelength, and (ii) smaller amplitude?

ANSWER (i) Wave (a) has the greater wavelength — it shows fewer waves (compressions are farther apart) over the same distance. (ii) Wave (a) also has the smaller amplitude — its density variation (height of the curve about the mean) is smaller than that of wave (b). So(a) is the lower-frequency, smaller-amplitude wave, while(b) is the higher-frequency, larger-amplitude wave.

6. The sound waves from three sources A, B and C are shown in Fig. 10.31. If the frequency of A is maximum and that of C is minimum, identify the curves and mark A, B and C.

ANSWER Frequency depends on the number of waves in a given distance (a shorter wavelength means a higher frequency), not on amplitude. A = the curve with the most compressions (shortest wavelength) — the green curve. C = the curve with the fewest compressions (longest wavelength) — the blue curve. B = the curve in between — the red curve.

7. Draw a graph of a sound wave with density amplitude 3 units and wavelength 4 cm.

ANSWER Plot density (y-axis) against distance (x-axis). Draw a smooth wave (sine-shaped) about the mean-density line. The curve should rise to +3 units above the mean and fall to 3 units below it (amplitude = 3 units), and one complete wave (crest + trough) should repeat every 4 cm along the distance axis.

8. In a movie, the explosion of a spacecraft in space is shown with a flash of light and sound at the same time. What are the errors in this depiction?

ANSWER Error 1: Space is a vacuum with no medium, so sound cannot travel there — no sound should be heard at all. Error 2: Even with a medium, light travels much faster than sound, so the flash would be seen well before the sound is heard — they cannot reach the viewer at the same time.

9. A source produces a sound wave of wavelength 3.44 m travelling at 344 m s-1. Find its time period.

ANSWER Frequency f = v / λ = 344 / 3.44 = 100 Hz. Time period T = 1 / f = 1 / 100 = 0.01 s.

10. A ship sends a sonar signal and detects an echo after 5 s. If the ultrasonic wave travels at 1525 m s-1 in seawater, how far down is the sunken ship?

ANSWER The signal travels down and back, so 2d = v × t = 1525 × 5 = 7625 m. Depth d = 7625 / 2 = 3812.5 m.

11. A parking sensor emits an ultrasonic wave that reflects off an obstacle. When the beep starts at 1.2 m from the obstacle, how much time does the wave take to travel to the obstacle and back? (Speed in air = 345 m s-1.)

ANSWER Total path = 2 × 1.2 = 2.4 m. Time t = distance / speed = 2.4 / 345 = 0.00696 s (about 7 milliseconds).

12. The speed of sound is about 331 m s-1 at 0 °C and 344 m s-1 at 22 °C. Roughly how much extra time will the sound of thunder take to travel 1720 m if the temperature changes from 22 °C to 0 °C?

ANSWER Time at 22 °C = 1720 / 344 = 5.00 s. Time at 0 °C = 1720 / 331 = 5.20 s. Extra time = 5.20 − 5.00 = about 0.2 s (sound travels slower in colder air).

13. The density variation of a sound wave travelling at 340 m s-1 is shown in Fig. 10.32. Calculate the wavelength and frequency.

ANSWER The 8 cm marked in the figure spans two complete waves (two compressions and two rarefactions), so the wavelength λ = 8 / 2 = 4 cm = 0.04 m. Frequency f = v / λ = 340 / 0.04 = 8500 Hz.

14. Two sound waves A and B (Fig. 10.33) travel at the same speed of 345 m s-1. Find the wavelength and frequency of each.

ANSWER From the graph, wave A completes one full wave in 2.5 cm, so λA = 2.5 cm = 0.025 m; wave B completes one full wave in 5.0 cm, so λB = 5.0 cm = 0.05 m. fA = v / λA = 345 / 0.025 = 13800 Hz. fB = v / λB = 345 / 0.05 = 6900 Hz.

15. Two identical sound sources at A (in air) and B (in water) send sound horizontally to a cliff and back (Fig. 10.34). If the time taken to return to A is 4.5 times that to B, what is the ratio of the speeds of sound in air and water?

ANSWER The horizontal distance is the same, so for each source 2d = v × t, i.e. t is inversely proportional to the speed. tA / tB = vwater / vair = 4.5. So vair : vwater = 1 : 4.5 (that is, 2 : 9) — sound travels much faster in water than in air.

Common Mistakes to Avoid

Watch out for these

  • Thinking sound can travel through vacuum — it cannot; it needs a medium.
  • Mixing up frequency (waves per second) with wavelength (distance per wave) — they are inversely related at fixed speed.
  • Forgetting the factor of 2 in echo/SONAR problems (the sound travels to the object and back).
  • Not converting cm to m before using v = f × λ.
  • Confusing echo (gap ≥ 0.1 s, distinct repeat) with reverberation (gap < 0.1 s, persistence).
  • Assuming amplitude affects frequency — loudness depends on amplitude, pitch depends on frequency.

Extra Practice Questions

Very Short Answer Type Questions

Q1. What type of wave is a sound wave?

ANSWERA longitudinal mechanical wave.

Q2. What is the SI unit of frequency?

ANSWERThe hertz (Hz).

Q3. On which property of a sound wave does its loudness depend?

ANSWERIts amplitude.

Short Answer Type Questions

Q1. Distinguish between the pitch and the loudness of a sound.

ANSWERPitch depends on the frequency — a higher frequency gives a higher pitch (shriller sound); loudness depends on the amplitude — a larger amplitude gives a louder sound.

Q2. A sound wave has a frequency of 500 Hz and a wavelength of 0.66 m. Calculate its speed.

ANSWERv = f × λ = 500 × 0.66 = 330 m s-1.

Long Answer Type Question

Q1. Explain how sound propagates through air using the idea of compressions and rarefactions.

ANSWER When an object vibrates, it pushes the nearby air particles together, forming a region of high density and pressure called a compression. As the object moves back, the air particles spread apart, forming a region of low density and pressure called a rarefaction. These compressions and rarefactions move forward one after another, carrying the sound energy through the air as a longitudinal wave, while the air particles only vibrate to and fro about their mean positions.

MCQs & Assertion–Reason

1. Sound waves are:

(a) transverse    (b) longitudinal    (c) electromagnetic    (d) stationary

2. Sound cannot travel through:

(a) solids    (b) liquids    (c) gases    (d) vacuum

3. The SI unit of frequency is:

(a) metre    (b) second    (c) hertz    (d) decibel

4. The relation between speed, frequency and wavelength is:

(a) v = f / λ    (b) v = f × λ    (c) v = λ / f    (d) v = f + λ

5. The minimum time gap to hear a distinct echo is:

(a) 0.01 s    (b) 0.1 s    (c) 1 s    (d) 10 s

6. The pitch of a sound depends on its:

(a) amplitude    (b) frequency    (c) speed    (d) wavelength only

7. SONAR is used to measure:

(a) air temperature    (b) depth of the sea    (c) light intensity    (d) wind speed

8. The frequency of ultrasound is:

(a) below 20 Hz    (b) 20 Hz to 20 000 Hz    (c) above 20 000 Hz    (d) exactly 20 Hz

9. If a sound wave has frequency 200 Hz, its time period is:

(a) 0.005 s    (b) 0.05 s    (c) 200 s    (d) 5 s

10. The loudness of a sound depends on its:

(a) frequency    (b) wavelength    (c) amplitude    (d) speed

Answer key: 1-(b), 2-(d), 3-(c), 4-(b), 5-(b), 6-(b), 7-(b), 8-(c), 9-(a), 10-(c).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: Sound cannot travel in a vacuum.

Reason: Sound is a mechanical wave that needs a medium to propagate.

A-R 2. Assertion: An echo is heard only if the reflecting surface is far enough away.

Reason: The reflected sound must reach the ear at least 0.1 s after the original.

A-R 3. Assertion: A higher-frequency sound has a higher pitch.

Reason: Pitch is determined by the amplitude of the wave.

A-R 4. Assertion: Sound travels faster in water than in air.

Reason: Particles are more closely packed in water, so vibrations pass on faster.

A-R 5. Assertion: SONAR uses ultrasound to find the depth of the sea.

Reason: Ultrasound is reflected by the seabed and the echo time is measured.

Answer key: 1-(A), 2-(A), 3-(C), 4-(A), 5-(A).

Quick Revision Summary

  • Sound is a longitudinal mechanical wave; it needs a medium and travels as compressions and rarefactions.
  • v = f × λ and T = 1/f.
  • Pitch depends on frequency; loudness depends on amplitude.
  • Echo: gap ≥ 0.1 s (distinct); reverberation: gap < 0.1 s (persistence).
  • In echo/SONAR problems use 2d = v × t (sound goes and returns).
  • Sound travels faster in solids and liquids than in gases.

Real-life Applications

Sound science is everywhere: SONAR measures sea depth and locates shipwrecks and shoals of fish; ultrasound scans are used in medicine and to clean and detect flaws in metals; bats and dolphins use echolocation to navigate; parking sensors use ultrasonic echoes; and concert halls are designed to control reverberation for clear sound.

How to score full marks in this chapter

Always write v = f × λ with consistent units (convert cm to m), and remember the factor of 2 for echoes and SONAR. Keep pitch (frequency) and loudness (amplitude) separate, and quote the 0.1 s rule when distinguishing an echo from reverberation.

Frequently Asked Questions

What is Class 9 Science Exploration Chapter 10 about?

Sound waves — how sound is produced and travels as a longitudinal wave, its characteristics (wavelength, frequency, amplitude, speed), v = f × λ, echo and reverberation, and applications such as SONAR and ultrasound.

What is the relation v = f × λ?

The speed of a wave equals its frequency multiplied by its wavelength; the time period is the reciprocal of the frequency (T = 1/f).

What is the difference between an echo and reverberation?

An echo is a distinct repeat heard when the reflected sound arrives at least 0.1 s later; reverberation is the persistence of sound from repeated reflections when the gap is shorter than 0.1 s.

Are these Class 9 Science Exploration Chapter 10 solutions free?

Yes. All solutions are free and follow the official NCERT Exploration textbook for 2026–27.

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