NCERT Solutions for Class 9th Science Chapter 10 End-of-chapter questions — Revise, Reflect, Refine

Book page 204 – 206 Updated on2026-09-08

Q1.

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

Answer

(ii) Sound needs a medium to propagate.

A mechanical wave is defined as a wave that requires a material medium to propagate. So the observation that pins sound down as mechanical is precisely the one that shows it dies out without a medium — the vacuum bell jar experiment.

Option	Is it true of sound?	Does it prove 'mechanical'?
(i) Shows reflection	Yes	No — light reflects too, and light is not mechanical
(ii) Needs a medium	Yes	Yes — this is the definition
(iii) Has frequency	Yes	No — every wave has a frequency
(iv) Carries energy	Yes	No — every wave carries energy

Why the other three fail: they are all true statements about sound, but they are true of all waves. An observation only counts as evidence for a classification if it distinguishes that class from the others.

Q2.

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

Answer

(iii) number of compressions per second.

Frequency is the number of complete density oscillations passing a fixed point each second, and every complete oscillation brings one compression with it. So raising the frequency directly raises the number of compressions arriving per second.

Speed: v depends only on the medium → unchanged
Wavelength: from v = ν × λ, λ = v/ν → decreases when ν increases
Time period: T = 1/ν → decreases when ν increases

Check with numbers (air, v = 344 m s⁻¹):
ν = 200 Hz → λ = 344 ÷ 200 = 1.72 m, T = 0.005 s
ν = 400 Hz → λ = 344 ÷ 400 = 0.86 m, T = 0.0025 s

Why the speed does not change: the speed of sound is fixed by the medium — how closely packed and how tightly bound its particles are, plus temperature and humidity. The source can only decide how often it sends compressions out. The medium then decides how fast each one travels, and the wavelength adjusts itself to fit.

Q3.

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

Answer

(ii) 5 Hz.

frequency ν = number of oscillations ÷ time taken
= 20 ÷ 4 s
= 5 s⁻¹ = 5 Hz

Each compression that sweeps past the point marks one complete density oscillation there, so counting compressions is the same as counting oscillations.

Time period T = 1/ν = 1 ÷ 5 Hz = 0.2 s — one compression every 0.2 s, which checks out: 20 × 0.2 s = 4 s ✓

Why 0.2 Hz is the trap: option (iv) is 4 ÷ 20, which is the time period in seconds, not the frequency. Watch the units — hertz means 'per second', so the count must go on top and the time underneath.

Q4.

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

Answer

It will produce reverberation, not an echo.

Time gap between the direct sound and the reflected sound = 0.05 s
Minimum gap needed to hear two sounds separately = 0.1 s

0.05 s < 0.1 s → the brain cannot separate them

Because the reflected sound arrives while the original is still being registered, the two merge. The sound seems to persist or linger a little after the source has stopped — and that persistence caused by reflections is exactly what reverberation means.

We can also check how big such a room is:

d = v t / 2 = (340 m s⁻¹ × 0.05 s) ÷ 2 = 17 m ÷ 2 = 8.5 m

A wall 8.5 m away is an ordinary large room — and indeed the minimum distance for a true echo is 17 m, twice as far.

	Echo	Reverberation
Time gap	At least 0.1 s	Less than 0.1 s (typically within 0.05 s)
Reflections involved	Usually a single reflection from a distant surface	Many reflections from the walls of a hall
What you hear	The sound repeated, clearly separate	The sound prolonged, blurred into itself
Where	Hills, cliffs, long empty corridors	Halls, auditoriums, empty rooms

Tip: auditoriums are designed with sound-absorbing panels, upholstered chairs and curtains so that reverberation is controlled. Too little and the hall sounds dead; too much and speech turns into a garble.

Q5.

Graphs representing two sound waves are given in Fig. 10.30. If the scales on the X and Y axes of the two graphs are the same, which of the two sound waves has (i) greater wavelength, and (ii) smaller amplitude?

Answer

(i) Wave (a) has the greater wavelength. (ii) Wave (a) also has the smaller amplitude.

Read the printed graphs across the same length of the distance axis:

	Wave (a)	Wave (b)
Complete cycles shown over the same distance	3	6
Wavelength (crest to crest)	Larger — about twice that of (b)	Smaller
Height of a crest above the average-density line	Smaller	Larger — roughly 1.5 times
So	Greater λ, smaller amplitude	Shorter λ, larger amplitude

Both graphs cover the same span of distance and use the same scales, so cycles can be counted directly and crest heights compared directly.

Why the scales had to be the same: wavelength is read off the horizontal axis and amplitude off the vertical axis. If the two graphs used different scales, a wave could look longer or taller without actually being so. Once the scales match, counting cycles and comparing crest heights is a fair comparison. In sound terms, (b) has the higher frequency (ν = v/λ) and, having the larger amplitude, also carries more energy — so it would be heard as a shriller and louder sound.

Q6.

The sound waves emitted by three sources A, B and C are represented in Fig. 10.31. If the frequency of A is maximum and C is minimum, identify the corresponding curves, and mark A, B and C on them.

Answer

The green curve is A, the red curve is B and the blue curve is C.

All three curves are drawn over the same distance axis, so the one that fits in the most cycles has the shortest wavelength and therefore the highest frequency.

Curve in Fig. 10.31	Complete cycles across the axis	Wavelength	Frequency ν = v/λ	Label
Green (tallest, most closely spaced)	4	Shortest	Maximum	A
Red (middle)	3	Middle	Middle	B
Blue (widest, smallest crests)	2	Longest	Minimum	C

All three travel in the same medium, so v is the same for all three.
From v = ν × λ, ν ∝ 1/λ — shorter wavelength means higher frequency.
Since λ_green < λ_red < λ_blue, ν_green > ν_red > ν_blue
∴ green = A, red = B, blue = C

Why you must not use the crest height: the tallest curve here also happens to be the one with the highest frequency, but that is a coincidence of the drawing. Crest height is amplitude, which tells you about loudness and energy, not about frequency. Frequency is read from how tightly packed the cycles are along the distance axis.

Tip: count from crest to crest, not from the start of the graph. If a curve begins mid-cycle, counting crests is far more reliable than trying to count whole cycles by eye.

Q7.

Draw a graph to represent a sound wave for which the density amplitude is 3 units and wavelength is 4 cm.

Answer

Plot density on the y-axis and distance on the x-axis, draw a horizontal dashed line for the average density, and then draw a smooth wave that

rises 3 units above the average-density line at each crest and falls 3 units below it at each trough (that is the density amplitude), and
repeats every 4 cm along the distance axis — crest to crest, or trough to trough, is 4 cm.

Density–distance graph with a density amplitude of 3 units and a wavelength of 4 cm. Each crest is a compression, each trough a rarefaction.

How to check your own drawing: (1) crest-to-crest and trough-to-trough must both measure 4 cm; (2) the crest must be exactly as far above the dashed line as the trough is below it, namely 3 units; (3) the curve must cross the average-density line at every quarter-wavelength, i.e. every 1 cm. Also mark C at each crest and R at each trough — that reminds you the graph is describing compressions and rarefactions.

Tip: if a speed were also given, you could add the frequency. For example in air at 344 m s⁻¹, ν = v/λ = 344 m s⁻¹ ÷ 0.04 m = 8600 Hz.

Q8.

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

Answer

There are two errors, and the first one is fatal.

There should be no sound at all. Outer space is a near vacuum. Sound is a mechanical wave and needs a material medium; with almost no particles there is nothing to compress and nothing to pass the disturbance on. The explosion would be completely silent to a distant observer. (Light is not a mechanical wave, so the flash would be seen — which is why sunlight reaches the Earth across empty space.)
Even in a medium, the two could not arrive together. Light travels enormously faster than sound, so the flash always arrives first and the sound follows after a delay.

speed of light = 3 × 10⁵ m s⁻¹ speed of sound in air = 340 m s⁻¹
ratio = 3 × 10⁵ ÷ 340 ≈ 880 000 times faster

For a source 1700 m away in air:
time for the sound = 1700 m ÷ 340 m s⁻¹ = 5 s
time for the light = 1700 m ÷ (3 × 10⁵ m s⁻¹) ≈ 0.0000057 s — effectively instant

The everyday version of the same physics: during a thunderstorm you see the lightning first and hear the thunder seconds later, even though both were produced at the same instant. Counting that delay and multiplying by 340 m s⁻¹ tells you how far away the strike was.

Tip: a fair correction for the film would be a bright silent flash, followed by nothing — or, if the camera is meant to be inside a spacecraft, the sound arriving through the ship's own metal hull, which is a medium.

Q9.

A source produces a sound wave of wavelength 3.44 m. If the wave travels with a speed of 344 m s–1 find its time period.

Answer

The time period is 0.01 s.

Step 1 — find the frequency. v = ν × λ (Eq. 10.2)
ν = v / λ = 344 m s⁻¹ ÷ 3.44 m
= 100 s⁻¹ = 100 Hz

Step 2 — find the time period. ν = 1/T (Eq. 10.1)
T = 1 / ν = 1 ÷ 100 Hz
= 0.01 s

Or in a single step, since v = λ/T: T = λ/v = 3.44 m ÷ 344 m s⁻¹ = 0.01 s.

What the answer means physically: T = 0.01 s is the time the wave takes to move forward by exactly one wavelength, and equally the time a single air particle takes to complete one full oscillation about its mean position. At 100 Hz this is a low, humming note, comfortably inside the audible range of 20 Hz – 20 kHz.

Tip: watch the units. m ÷ (m s⁻¹) = s, so dividing a wavelength by a speed always gives a time. If your answer comes out in the wrong unit, you have divided the wrong way round.

Q10.

A ship searching for a sunken ship sent a sonar signal and detected an echo after 5 s. If ultrasonic wave travels at 1525 m s–1 in seawater, approximately how far down in the ocean is the wreckage of the sunken ship located?

Answer

The wreck lies about 3812.5 m (roughly 3.8 km) below the ship.

The 5 s is the time for the pulse to go down to the wreck and back.

total distance = speed × time
= 1525 m s⁻¹ × 5 s
= 7625 m

depth of the wreck = total distance ÷ 2
= 7625 m ÷ 2
= 3812.5 m ≈ 3.8 km

Why ultrasound is used here: ultrasonic waves have very short wavelengths, so they reflect sharply from an object as small as a ship's hull instead of spreading around it, and they can be sent out as a narrow beam that gives a definite direction. Sound also travels far in water without dying out, unlike light. That combination — direction, sharp reflection and long reach — is what makes sonar work.

Tip: the number 1525 m s⁻¹ is close to the 1500 m s⁻¹ in Table 10.1. Speed in seawater varies a little with temperature, depth and salinity, so a question will always tell you which value to use.

Q11.

A vehicle is fitted with an ultrasonic distance sensor as part of parking assistance system which provides echolocation, while the driver is reversing the vehicle. It emits ultrasonic wave (about 40 kHz) which is reflected by the obstacle. When the warning beep starts sounding at a distance of 1.2 m from the obstacle, how much time is taken by ultrasonic wave to travel to the obstacle and come back? Assume the speed of ultrasonic wave in air to be 345 m s–1.

Answer

About 6.96 × 10⁻³ s, that is roughly 7 milliseconds.

The pulse goes to the obstacle and returns, so it covers twice 1.2 m.

total distance = 2 × 1.2 m = 2.4 m

time = distance ÷ speed
= 2.4 m ÷ 345 m s⁻¹
= 0.00696 s ≈ 6.96 × 10⁻³ s ≈ 7.0 ms

We can also confirm that 40 kHz really is ultrasonic and see why it suits the job:

λ = v / ν = 345 m s⁻¹ ÷ 40 000 s⁻¹ = 0.0086 m ≈ 0.9 cm
40 kHz > 20 kHz → ultrasonic, inaudible to the driver

Why the system works so well: the round trip takes only about 7 ms, so the sensor can fire many pulses every second and update the beep almost continuously as the car creeps back. The wavelength of under a centimetre means the pulse reflects cleanly off a low bollard or a kerb. And because 40 kHz is above the audible range, the driver hears only the electronic beep, never the ultrasound itself.

Tip: milliseconds are natural units here. 0.00696 s = 6.96 ms, since 1 ms = 10⁻³ s.

Q12.

The speed of sound in air is about 331 m s–1 at 0 ºC and nearly 344 m s–1 at 22 ºC. Roughly how much extra time will the sound of thunder take to travel a distance of 1720 m, if the air temperature changes from 22 ºC to 0 ºC? Assume that all other conditions remain unchanged.

Answer

About 0.2 s of extra time.

time = distance ÷ speed

At 22 ºC (v = 344 m s⁻¹):
t₁ = 1720 m ÷ 344 m s⁻¹ = 5.000 s

At 0 ºC (v = 331 m s⁻¹):
t₂ = 1720 m ÷ 331 m s⁻¹ = 5.196 s

Extra time = t₂ − t₁
= 5.196 s − 5.000 s = 0.196 s ≈ 0.2 s

Why cold air slows sound down: in a gas, sound travels by molecules colliding with their neighbours. Temperature is a measure of how fast those molecules are already moving about. At 22 ºC the molecules are moving faster than at 0 ºC, so they meet their neighbours sooner and pass the compression on more quickly. Cool the air and every collision is slightly delayed, so the whole disturbance creeps forward more slowly. Humidity works the same way — moister air carries sound slightly faster.

Tip: notice how small the effect is: a 22 ºC drop changes a 5 s journey by only about 4%. That is why the rough rule for a thunderstorm — 'count the seconds and multiply by about 340 m s⁻¹' — works well enough on a winter night as on a summer one.

Q13.

The variation of density of medium for a sound wave propagating with a speed of 340 m s–1 is shown in Fig. 10.32. Calculate the wavelength and frequency of the sound wave.

Answer

Wavelength λ = 4 cm = 0.04 m, and frequency ν = 8500 Hz.

First read the wavelength off the figure. The dot pattern shows alternating dense bands (compressions) and thin bands (rarefactions), with faint guide lines drawn through their centres. The arrow marked 8 cm stretches across two complete repeats of the pattern — that is, from one compression to the third compression. Therefore

8 cm = 2λ → λ = 8 cm ÷ 2 = 4 cm = 0.04 m

Reading Fig. 10.32: the 8 cm arrow spans two whole repeats of the compression pattern, so one wavelength is 4 cm.

Now the frequency. v = ν × λ (Eq. 10.2)
ν = v / λ
= 340 m s⁻¹ ÷ 0.04 m
= 8500 s⁻¹ = 8500 Hz

Why the conversion to metres matters: the speed is given in m s⁻¹, so the wavelength must be in metres before dividing — 4 cm = 4/100 m = 0.04 m. Working in centimetres would give 340 ÷ 4 = 85, which is not a frequency in hertz at all. Always bring every quantity to SI units before substituting.

Check it yourself: 8500 Hz lies between 20 Hz and 20 kHz, so this is an audible sound — a high, shrill note. The time period is T = 1/ν = 1/8500 ≈ 1.18 × 10⁻⁴ s.

Q14.

The graphical representation of two sound waves A and B propagating at the same speed of 345 m s–1 is shown in Fig. 10.33. What is the wavelength of each of them? Also, calculate their frequencies.

Answer

λ_A = 2.5 cm and ν_A = 13 800 Hz; λ_B = 5.0 cm and ν_B = 6900 Hz.

Read the wavelengths off the distance axis of Fig. 10.33. Between 0 and 5.0 cm, curve A completes two full cycles while curve B completes just one.

Wave A: 2 wavelengths in 5.0 cm → λ_A = 5.0 cm ÷ 2 = 2.5 cm = 0.025 m
Wave B: 1 wavelength in 5.0 cm → λ_B = 5.0 cm = 0.050 m

Frequencies, using ν = v / λ with v = 345 m s⁻¹ for both:

ν_A = 345 m s⁻¹ ÷ 0.025 m = 13 800 Hz = 13.8 kHz
ν_B = 345 m s⁻¹ ÷ 0.050 m = 6900 Hz = 6.9 kHz

Wave	Cycles in 5.0 cm	Wavelength λ	Frequency ν = v/λ	Time period T = 1/ν
A	2	2.5 cm = 0.025 m	13 800 Hz	7.25 × 10⁻⁵ s
B	1	5.0 cm = 0.050 m	6900 Hz	1.45 × 10⁻⁴ s

Why the frequencies come out in the ratio 2 : 1: both waves are in the same medium, so both travel at 345 m s⁻¹. With v fixed, ν is inversely proportional to λ. Wave A's wavelength is exactly half of B's, so its frequency is exactly double. In musical language A is one octave above B — an octave is precisely a doubling of frequency.

Tip: the two curves also differ in amplitude (A's crests are taller), which affects loudness. The question asks only about wavelength and frequency, so amplitude plays no part in the calculation.

Q15.

Two identical sound sources are placed at A and B — one in air and one submerged in water (Fig. 10.34). Both produce sounds at the same time, which travel horizontally to the vertical side of the cliff and come back. If the time taken by the sound to return to A is 4.5 times than that of B, what is the ratio between the speeds of sound in air and water?

Answer

v_air : v_water = 1 : 4.5 = 2 : 9, i.e. about 0.22 : 1.

Source A is in air and source B is submerged in water, but both are the same horizontal distance from the cliff face, so both pulses cover the same round-trip distance. Call that distance 2d for each.

time = distance ÷ speed

For A (in air): t_A = 2d / v_air
For B (in water): t_B = 2d / v_water

Given: t_A = 4.5 × t_B

∴ 2d / v_air = 4.5 × (2d / v_water)
The 2d cancels from both sides:
1 / v_air = 4.5 / v_water
∴ v_water = 4.5 × v_air

v_air : v_water = 1 : 4.5 = 2 : 9 ≈ 0.22

Both pulses cover the same round trip 2d to the cliff and back. Only the speed of the medium differs, so the times are in inverse ratio to the speeds.

Why the answer is physically sensible: the chapter states that sound travels about 4 – 5 times faster in water than in air, and this problem gives exactly 4.5. Using the values in Table 10.1, 1500 m s⁻¹ ÷ 340 m s⁻¹ = 4.4 — a very close match. Water's molecules are far more closely packed than air's, so a compression is passed on much more quickly.

Tip: be careful which way round the ratio is asked. The question asks for air : water, which is the smaller number first (1 : 4.5). If it had asked for water : air, the answer would be 4.5 : 1.

Note on the printed figure: in the English edition Fig. 10.34 is printed with only the labels A and B and no picture. The Hindi edition prints the full scene, which shows source A mounted in the air on a pier and source B hanging below the water surface, both facing the vertical rock face across the water. The physics is unaffected: A and B are at the same horizontal distance from the cliff.

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