NCERT Class 9 Science Chapter 10 Solutions – Sound Waves
Looking for accurate and complete ncert class 9 science chapter 10 solutions for the new CBSE curriculum (2026-27)? You are in the right place! In this guide, we provide step-by-step, direct, and easy-to-understand solutions for all exercise questions from NCERT Exploration Class 9 Science Chapter 10: Sound Waves: Characteristics and Applications.
These answers are crafted specifically to help students write clean, exam-ready answers in their notebooks without extra confusion or lengthy fluff.
Question 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
Correct Option: (ii) Sound needs a medium to propagate
Explanation: Mechanical waves require a material medium (solid, liquid, or gas) consisting of particles to undergo compression and rarefaction for energy transfer. The inability of sound to travel through a vacuum proves it is a mechanical wave.
Question 2
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
Correct Option: (iii) number of compressions per second
Explanation: Frequency is defined as the number of oscillations or compressions passing a given point per second ($f = \frac{1}{T}$).
Question 3
If 20 compressions pass a point in 4 seconds, the frequency is:
(i) $80\text{ Hz}$ (ii) $5\text{ Hz}$ (iii) $10\text{ Hz}$ (iv) $0.2\text{ Hz}$
Correct Option: (ii) 5 Hz
Calculation: $$\text{Frequency } (f) = \frac{\text{Number of compressions}}{\text{Time}} = \frac{20}{4} = \mathbf{5\text{ Hz}}$$
Question 4
In a room, the reflected sound reaches the ear $0.05\text{ s}$ after its production. Will it produce an echo or reverberation? Justify your answer.
It will produce an echo.
Justification: The persistence of human hearing for sound is about $0.1\text{ s}$. For a distinct echo to be heard, the time interval between original and reflected sound must be at least $0.1\text{ s}$. Since $0.05\text{ s}$ is less than $0.1\text{ s}$, the sound overlaps with the original sound, resulting in reverberation (or overlapping persistence, though strictly time $<0.1\text{ s}$ causes reverberation rather than a distinct standalone echo).
Question 5
Graphs representing two sound waves are given in Fig. 10.30. If scales on X and Y axes are the same, which wave has (i) greater wavelength, and (ii) smaller amplitude?
- (i) Greater wavelength: Wave (a) has fewer cycles per unit distance, meaning its wavelength ($\lambda$) is larger.
- (ii) Smaller amplitude: Wave (a) has a smaller peak height (density variation) from the mean line compared to Wave (b).
Question 6
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.
- Source A (Maximum frequency): Corresponds to the wave with the highest number of compressions/waves packed closely together (shortest wavelength).
- Source C (Minimum frequency): Corresponds to the wave with the fewest cycles (longest wavelength).
- Source B: Corresponds to the intermediate wave curve.
Question 7
Draw a graph to represent a sound wave for which the density amplitude is 3 units and wavelength is 4 cm.
To sketch this in your notebook:
- Draw density on the Y-axis with peaks reaching $+3$ units and troughs at $-3$ units from the normal density line.
- Draw distance on the X-axis such that one complete wave cycle (crest + trough) spans exactly $4\text{ cm}$.
Question 8
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?
Two major scientific errors:
- Sound in space: Space is a vacuum lacking any material medium. Sound waves cannot travel through a vacuum, so an explosion in space would be completely silent.
- Simultaneous arrival: Even if sound could travel through space, light travels much faster ($c = 3 \times 10^8\text{ m s}^{-1}$) than sound, so the flash of light would always be seen long before any sound wave arrived.
Question 9
A source produces a sound wave of wavelength $3.44\text{ m}$. If the wave travels with a speed of $344\text{ m s}^{-1}$, find its time period.
- Wavelength ($\lambda$) = $3.44\text{ m}$
- Speed ($v$) = $344\text{ m s}^{-1}$
- Frequency ($f$): $$f = \frac{v}{\lambda} = \frac{344}{3.44} = 100\text{ Hz}$$
- Time Period ($T$): $$T = \frac{1}{f} = \frac{1}{100} = \mathbf{0.01\text{ s}}$$
Question 10
A ship searching for a sunken ship sent a sonar signal and detected an echo after $5\text{ s}$. If ultrasonic wave travels at $1525\text{ m s}^{-1}$ in seawater, how far down is the wreckage?
- Total echo time ($t$) = $5\text{ s}$ (round trip: down and back up).
- One-way travel time ($t_{\text{one-way}}$) = $\frac{5}{2} = 2.5\text{ s}$
- Speed of sound in water ($v$) = $1525\text{ m s}^{-1}$
- Depth of wreckage ($d$): $$d = v \times t_{\text{one-way}} = 1525 \times 2.5 = \mathbf{3812.5\text{ m} \text{ (or } 3.81\text{ km)}$$
Question 11
A vehicle is fitted with an ultrasonic distance sensor emitting waves at $\approx 40\text{ kHz}$. When warning beeps start at a distance of $1.2\text{ m}$ from an obstacle, how much time is taken by the ultrasonic wave to travel to the obstacle and come back? ($v = 345\text{ m s}^{-1}$).
- Distance to obstacle ($d$) = $1.2\text{ m}$
- Total distance for round trip ($2d$) = $2 \times 1.2 = 2.4\text{ m}$
- Speed ($v$) = $345\text{ m s}^{-1}$
- Time taken ($t$): $$t = \frac{\text{Total Distance}}{\text{Speed}} = \frac{2.4}{345} \approx \mathbf{0.00696\text{ s} \text{ (or } 6.96\text{ ms})}$$
Question 12
The speed of sound in air is $331\text{ m s}^{-1}$ at $0\text{ }^\circ\text{C}$ and $344\text{ m s}^{-1}$ at $22\text{ }^\circ\text{C}$. Roughly how much extra time will the sound of thunder take to travel $1720\text{ m}$ if the air temperature drops from $22\text{ }^\circ\text{C}$ to $0\text{ }^\circ\text{C}$?
- Distance ($d$) = $1720\text{ m}$
- Time at $22^\circ\text{C}$ ($t_{22}$): $$t_{22} = \frac{1720}{344} = 5.0\text{ s}$$
- Time at $0^\circ\text{C}$ ($t_{0}$): $$t_{0} = \frac{1720}{331} \approx 5.196\text{ s}$$
- Extra Time Taken: $$\Delta t = 5.196\text{ s} - 5.0\text{ s} = \mathbf{0.196\text{ s}}$$
Question 13
The variation of density of medium for a sound wave propagating with a speed of $340\text{ m s}^{-1}$ is shown in Fig. 10.32 ($8\text{ cm}$ span). Calculate wavelength and frequency.
- From Fig. 10.32, $2$ complete wavelengths span across $8\text{ cm}$.
- Wavelength ($\lambda$): $$\lambda = \frac{8\text{ cm}}{2} = 4\text{ cm} = \mathbf{0.04\text{ m}}$$
- Frequency ($f$): $$f = \frac{v}{\lambda} = \frac{340\text{ m s}^{-1}}{0.04\text{ m}} = \mathbf{8500\text{ Hz} \text{ (or } 8.5\text{ kHz})}$$
Question 14
The graphical representation of two sound waves A and B propagating at $345\text{ m s}^{-1}$ is shown in Fig. 10.33 ($5.0\text{ cm}$ span). What is the wavelength and frequency of each?
- Wave A: Completes $3$ full waves in $5.0\text{ cm}$.
- $$\lambda_A = \frac{5.0\text{ cm}}{3} = 1.67\text{ cm} = 0.0167\text{ m}$$
- $$f_A = \frac{v}{\lambda_A} = \frac{345}{0.0167} \approx \mathbf{20,700\text{ Hz} \text{ (or } 20.7\text{ kHz})}$$
- Wave B: Completes $1$ full wave in $5.0\text{ cm}$.
- $$\lambda_B = \frac{5.0\text{ cm}}{1} = 5.0\text{ cm} = 0.05\text{ m}$$
- $$f_B = \frac{v}{\lambda_B} = \frac{345}{0.05} = \mathbf{6900\text{ Hz} \text{ (or } 6.9\text{ kHz})}$$
Question 15
Two identical sound sources are placed at A (in air) and B (submerged in water) (Fig. 10.34). Both produce sounds traveling to a cliff and back. If time taken by sound in air is $4.5$ times that in water, what is the ratio between speeds of sound in air and water?
- Let distance to cliff be $d$ (round trip distance $2d$).
- Time in air ($t_{\text{air}}$) = $4.5 \times t_{\text{water}}$
- $$\frac{2d}{v_{\text{air}}} = 4.5 \times \frac{2d}{v_{\text{water}}} \implies v_{\text{water}} = 4.5 \times v_{\text{air}}$$
- Ratio of speed in air to speed in water ($\frac{v_{\text{air}}}{v_{\text{water}}}$): $$\frac{v_{\text{air}}}{v_{\text{water}}} = \frac{1}{4.5} = \mathbf{\frac{2}{9} \text{ (or } 0.222:1)}$$
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Based on latest NCERT class 9 book.