NCERT Solutions Class 9 Science Ch 10 Sound 2026-27
NCERT Solutions Class 9 Science Chapter 10 Sound Waves 2026-27. All 33 questions solved - wavelength, frequency, echo, sonar.
This free NCERT Solutions for CBSE Class IX Science, Chapter 10: Sound Waves: Characteristics and Applications, contains step-by-step solutions to every NCERT textbook question and exercise. It has been prepared by Sumeet Sahu at Unique Study Point, Indore, strictly following the latest NCERT syllabus for Session 2026-27.
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- First revise the chapter — Sound Waves: Characteristics and Applications from your notes or textbook.
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These are the NCERT Solutions for Class 9 Science Chapter 10, Sound Waves: Characteristics and Applications, from the new NCERT Science textbook Exploration (session 2026-27). All 33 questions are solved step by step, covering how sound travels, wavelength and frequency, echo and SONAR, and graphs of sound waves.
- Class
- 9 (IX)
- Subject
- Science
- Textbook
- Exploration (NCERT, 2026-27)
- Chapter
- 10 – Sound Waves: Characteristics and Applications
- Questions solved
- 33
Jump to a question (33 questions)
NCERT Class 9 Science Chapter 10 Questions and Answers
Question 1: Can astronauts hear each other in space?
Two astronauts are repairing the arm of a space station together during a spacewalk. Can they talk to each other and hear the sounds of metal clanking as they do on the Earth?
No. Sound needs a material medium (air, water or a solid) to travel. Space is a vacuum, so there is no medium, and sound waves cannot travel. So the astronauts cannot talk to each other directly or hear the metal clanking, even if the sound is produced.
Question 2: Ways of producing sound
Explore various ways of producing sound.
Sound is produced when an object vibrates. Some ways are:
- Plucking: plucking a string (guitar, sitar).
- Striking: striking an object (drum, bell).
- Blowing: the air column vibrates (flute, whistle).
- Rubbing (friction): rubbing surfaces (a violin bow on strings).
In every case the sound comes from the vibrations of an object or of an air column.
Question 3: Musical instruments and their vibrating parts
Make a list of different types of musical instruments and identify their vibrating parts which produce sound.
| Type of instrument | Examples | Vibrating part |
|---|---|---|
| String instruments | Guitar, sitar, violin | Stretched strings |
| Wind instruments | Flute, trumpet, clarinet | Air column inside the instrument |
| Membrane instruments | Drum, tabla | Stretched membrane (skin) |
| Percussion (solid) instruments | Bell, cymbals | The whole body of the instrument |
| Keyboard instruments | Piano | Stretched strings inside, struck by hammers |
Question 4: How bats locate their prey
How do most bats use sound to locate their prey in the dark at night?
Bats use echolocation. They produce ultrasonic sound waves (high-frequency sounds that humans cannot hear). These waves hit objects such as insects and are reflected back as echoes. By detecting the echoes, the bat finds the position, the distance and even the size of its prey.
Question 5: Assertion-reason: bell in a jar
Assertion (A): We cannot hear the sound of a bell ringing in a closed jar after most of the air is pumped out.
Reason (R): Sound requires a medium to travel.
- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
Correct option: (a)
Sound cannot travel through a vacuum. When the air is removed from the jar there is no medium for the sound, so the sound of the bell cannot be heard. R explains A.
Question 6: Assertion-reason: compressions and rarefactions
Assertion (A): Compressions and rarefactions move through the medium.
Reason (R): Individual particles of the medium continuously move forward with the wave.
- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
Correct option: (c)
- A is true: compressions and rarefactions travel through the medium as a sound wave moves.
- R is false: the particles do not move forward with the wave. They only oscillate back and forth about their mean position.
Question 7: What actually reaches your ear from a tuning fork?
When sound travels from a tuning fork to your ear, which of the following actually reaches your ear?
- Air particles near the tuning fork
- A continuous stream of compressed air
- The tuning fork material
- Energy carried by sound waves
Correct option: (d) Energy carried by sound waves
The air particles only vibrate back and forth and do not travel from the source to the ear. What is transferred is the energy of the wave, not matter.
Question 8: Oscillations of a piston per minute
If the frequency of a sound wave produced by an oscillating piston of a long tube filled with air is 20 Hz, then how many oscillations does the piston complete per minute?
- Meaning of 20 HzThe piston completes 20 oscillations in every second.
- In 1 minute (60 s)20 × 60 = 1200
The piston completes 1200 oscillations per minute.
Question 9: Half of the wavelength from a density-distance graph
For the sound wave represented by the graph shown in the figure, what is half of its wavelength?
One wavelength is the distance between two consecutive compressions (peaks) or two consecutive rarefactions (troughs). From the graph, this distance is 3.0 cm.
λ = 3.0 cm ⇒ λ ÷ 2 = 1.5 cmHalf of the wavelength = 1.5 cm
Question 10: Comparing the speed of sound in different media
The table shows the speed of sound in a few media at atmospheric pressure. Compare the speeds in different media by finding the ratio of (i) the speed of sound in water with respect to the speed in air, (ii) the speed of sound in steel with respect to the speed in water.
| State | Substance | Approximate speed |
|---|---|---|
| Solid | Steel | 5000 m s−1 |
| Liquid | Water | 1500 m s−1 |
| Gas | Air | 340 m s−1 |
- (i) Water to air1500 ÷ 340 = 75 ÷ 17 ≈ 4.41
- (ii) Steel to water5000 ÷ 1500 = 10 ÷ 3 ≈ 3.33
(i) 75 : 17 (about 4.4 : 1) (ii) 10 : 3 (about 3.3 : 1)
Question 11: Minimum distance of a reflecting surface for an echo
An experiment is being set up that requires echoes to arrive at least 0.2 s after the emission of sound. What minimum distance should a reflecting surface be placed at? Assume the speed of sound to be 343 m s−1.
- Total distance covered by the soundThe sound goes to the surface and comes back.2d = v × t = 343 × 0.2 = 68.6 m
- Distance of the surfaced = 68.6 ÷ 2 = 34.3 m
The reflecting surface should be at least 34.3 m away.
Question 12: Depth of the ocean from a sonar signal
A sonar signal sent to find the depth of ocean takes 4 s to return. What is the depth of the ocean at that location if the speed of sound in seawater is 1500 m s−1?
- Total distancev × t = 1500 × 4 = 6000 m
- Depth (one way)6000 ÷ 2 = 3000 m
The depth of the ocean is 3000 m.
Question 13: Frequency and time period from the number of oscillations
If there are 10 density oscillations in 2 seconds at a given position, then calculate the (i) frequency of the sound wave, and (ii) its time period.
- (i) Frequencyf = number of oscillations ÷ time = 10 ÷ 2 = 5 Hz
- (ii) Time periodT = 1 ÷ f = 1 ÷ 5 = 0.2 s
f = 5 Hz and T = 0.2 s
Question 14: Wavelengths for the limits of human hearing
Human hearing roughly spans 20 Hz to 20 kHz. What are the corresponding wavelengths in air for these two frequencies? Use the speed of sound in air as 344 m s−1.
- Relationv = νλ, so λ = v ÷ ν
- For 20 Hzλ = 344 ÷ 20 = 17.2 m
- For 20 kHz = 20000 Hzλ = 344 ÷ 20000 = 0.0172 m = 1.72 cm
17.2 m and 1.72 cm
Question 15: Distance of a lightning strike
During a thunderstorm, lightning is seen before thunder is heard because sound travels much slower than light. If the time delay between seeing the lightning flash and hearing the thunder is measured to be 5 s, estimate the distance to the lightning strike. Use the speed of sound in air as 340 m s−1. Assume that light reaches you almost instantaneously.
The lightning struck about 1.7 km away.
Question 16: Wavelength, frequency and time period of a wave in steel
From the graphical representation of a sound wave propagating in steel, find its wavelength. Calculate its frequency and time period if the speed of sound in steel is 5000 m s−1.
- WavelengthFrom the graph, the distance between two consecutive peaks is 50 m.λ = 50 m
- Frequencyν = v ÷ λ = 5000 ÷ 50 = 100 Hz
- Time periodT = 1 ÷ ν = 1 ÷ 100 = 0.01 s
λ = 50 m, frequency = 100 Hz, time period = 0.01 s
Question 17: Distance from the wall after a clap
You clap in an empty corridor and hear an echo after 0.5 s. If the speed of sound in air is 340 m s−1, calculate your distance from the wall.
The wall is 85 m away.
Question 18: Distance of an object from a naval sonar signal
A naval sonar signal sent into seawater returns after 0.90 s. The speed of sound in seawater is 1530 m s−1. How far is the object?
- Time to reach the object0.90 ÷ 2 = 0.45 s
- Distance1530 × 0.45 = 688.5 m
The object is 688.5 m away.
Question 19: Observation supporting that sound is a mechanical wave
Which observation best supports the idea that sound is a mechanical wave?
- Sound shows reflection
- Sound carries energy
- Sound needs a medium.
- Sound has frequency
Correct option: (c) Sound needs a medium.
Mechanical waves need a material medium to travel. Sound cannot travel in a vacuum, which shows it is mechanical. Reflection, energy and frequency are properties of many waves, including light, which is not mechanical.
Question 20: Effect of increasing the frequency of a wave
For a sound wave propagating in a medium, increasing its frequency will increase its
- Increases compressions per second.
- wavelength
- speed
- time period
Correct option: (a) Increases compressions per second.
Frequency is the number of compressions and rarefactions per second. If it increases, the wavelength decreases, the time period decreases, and the speed stays constant in the same medium.
Question 21: Frequency from the number of compressions
If 20 compressions pass a point in 4 seconds, the frequency is
- 80 Hz
- 0.2 Hz
- 10 Hz
- 5 Hz
Correct option: (d) 5 Hz
f = number of compressions ÷ time = 20 ÷ 4 = 5 HzQuestion 22: Echo or reverberation after 0.05 s?
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.
The sensation of a sound stays in our brain for about 0.1 s (persistence of hearing).
- Condition for an echo: the reflected sound must reach the ear at least 0.1 s after the original sound.
- Here: the reflected sound arrives after only 0.05 s, which is less than 0.1 s.
So the original and the reflected sounds overlap and are heard as one prolonged sound. This is reverberation, not an echo.
Question 23: Wavelength and amplitude of two waves
Graphs representing two sound waves are given in the figure. 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?
(i) Greater wavelength: wave (a). The wavelength is the distance between two consecutive peaks. The peaks in graph (a) are much farther apart than those in graph (b), and the scales are the same.
(ii) Smaller amplitude: wave (a). The amplitude is the maximum height of the wave from the middle (dashed) line. The peaks of wave (a) are shorter than those of wave (b). So wave (a) is softer and wave (b) is louder.
Question 24: Identifying sources A, B and C from the waves
The sound waves emitted by three sources A, B and C are represented in the figure. If the frequency of A is maximum and C is minimum, identify the corresponding curves, and mark A, B and C on them.
The wave with the most oscillations (the shortest wavelength) has the highest frequency, and the wave with the fewest oscillations (the longest wavelength) has the lowest frequency.
| Source | Curve | Reason |
|---|---|---|
| A | Green curve | Most oscillations, so maximum frequency |
| B | Red curve | In between |
| C | Blue curve | Fewest oscillations, so minimum frequency |
Question 25: Graph of a sound wave with amplitude 3 units and wavelength 4 cm
Draw a graph to represent a sound wave for which the density amplitude is 3 units and wavelength is 4 cm.
Draw the density on the y-axis and the distance (in cm) on the x-axis, with a dashed line in the middle for the normal density.
- The peaks (compressions) reach +3 units and the troughs (rarefactions) reach −3 units, so the amplitude is 3 units.
- The distance between two consecutive peaks is 4 cm, so the wavelength is 4 cm. The crests come at 1 cm, 5 cm, 9 cm and so on.
Question 26: Errors in a movie scene of a spacecraft explosion
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?
- No sound in space: space is a vacuum, so sound cannot travel. Only the light of the explosion could be seen, not heard.
- Not at the same time: even if sound could travel, light moves much faster than sound, so the two could not be observed at the same time.
Question 27: Time period of a wave of wavelength 3.44 m
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.
- Frequencyν = v ÷ λ = 344 ÷ 3.44 = 100 Hz
- Time periodT = 1 ÷ ν = 1 ÷ 100 = 0.01 s
The time period is 0.01 s.
Question 28: Depth of a sunken ship from a sonar echo
A ship searching for a sunken ship sent a sonar signal and detected an echo after 5 s. If the 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?
- Total distance1525 × 5 = 7625 m
- Depth (one way)7625 ÷ 2 = 3812.5 m
The wreckage is about 3812.5 m deep.
Question 29: Time taken by an ultrasonic parking sensor
A vehicle is fitted with an ultrasonic distance sensor as part of a parking assistance system. It emits an 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 the ultrasonic wave to travel to the obstacle and come back? Assume the speed of the ultrasonic wave in air to be 345 m s−1.
- Total distanceThe wave goes to the obstacle and back.2d = 2 × 1.2 = 2.4 m
- Timet = 2d ÷ v = 2.4 ÷ 345 ≈ 0.007 s
The time taken is about 0.007 s (7 ms).
Question 30: Extra time taken by thunder when the temperature falls
The speed of sound in air is about 331 m s−1 at 0°C and nearly 344 m s−1 at 22°C. 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°C to 0°C? Assume that all other conditions remain unchanged.
- Time at 22°Ct1 = 1720 ÷ 344 = 5 s
- Time at 0°Ct2 = 1720 ÷ 331 ≈ 5.2 s
- Extra timeΔt = t2 − t1 ≈ 5.2 − 5 = 0.2 s
The sound takes about 0.2 s more time.
Question 31: Wavelength and frequency from the density variation of a medium
The variation of density of the medium for a sound wave propagating with a speed of 340 m s−1 is shown in the figure. Calculate the wavelength and frequency of the sound wave.
- WavelengthThe 8 cm marked in the figure runs from the first dense region (compression) to the third one, which covers two wavelengths.2λ = 8 cm ⇒ λ = 4 cm = 0.04 m
- Frequencyf = v ÷ λ = 340 ÷ 0.04 = 8500 Hz
Wavelength = 0.04 m (4 cm) and frequency = 8500 Hz
Question 32: Wavelengths and frequencies of waves A and B
The graphical representation of two sound waves A and B propagating at the same speed of 345 m s−1 is shown in the figure. What is the wavelength of each of them? Also, calculate their frequencies.
- Wavelength of AWave A starts at 0 and one full oscillation ends at 2.5 cm.λA = 2.5 cm = 0.025 m
- Wavelength of BWave B crosses the middle line at 2.5 cm (half a wavelength) and completes one oscillation at 5.0 cm.λB = 5.0 cm = 0.05 m
- FrequenciesfA = v ÷ λA = 345 ÷ 0.025 = 13,800 HzfB = v ÷ λB = 345 ÷ 0.05 = 6,900 Hz
A: 2.5 cm and 13,800 Hz B: 5 cm and 6,900 Hz
Question 33: Ratio of the speeds of sound in air and water
Two identical sound sources are placed at A and B, one in air and one submerged in water. 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 that of B, what is the ratio between the speeds of sound in air and water?
- For the same distance dThe sound travels 2d, so the time t = 2d ÷ v. For the same distance, the time is inversely proportional to the speed.
- TimestA = 2d ÷ vair tB = 2d ÷ vwater
- Given tA = 4.5 tB1 ÷ vair = 4.5 × (1 ÷ vwater) ⇒ vair ÷ vwater = 1 ÷ 4.5 = 2 ÷ 9
vair : vwater = 2 : 9
Quick revision: Class 9 Science Chapter 10
- Sound is produced by vibrations and needs a material medium; it cannot travel through a vacuum.
- Sound is a longitudinal mechanical wave: compressions and rarefactions travel through the medium, but the particles only oscillate about their mean positions.
- v = νλ, frequency ν = 1 ÷ T; the speed of sound is greatest in solids and least in gases, and it increases with temperature.
- Amplitude decides the loudness and frequency decides the pitch of a sound.
- An echo is heard when the reflected sound arrives at least 0.1 s after the original; for a distance d, 2d = v × t.
- Reverberation is the prolonged sound due to repeated reflections when the delay is less than 0.1 s.
- Humans hear from 20 Hz to 20,000 Hz; sounds above 20,000 Hz are ultrasonic and are used in SONAR and by bats.
- Light travels much faster than sound, so lightning is seen before thunder is heard.
Frequently asked questions
How many questions are solved in Class 9 Science Chapter 10?
All 33 questions of Chapter 10, Sound Waves: Characteristics and Applications, from the new NCERT Exploration textbook (2026-27) are solved on this page.
What is the difference between an echo and reverberation?
An echo is a distinct repetition of sound heard when the reflected sound arrives at least 0.1 s after the original. If it arrives sooner, the sounds overlap and are heard as a prolonged sound called reverberation.
How do you find the depth of the sea using SONAR?
The signal travels to the bottom and back, so depth = (speed of sound × time taken) ÷ 2.
What is the relation between speed, frequency and wavelength?
The speed of a wave equals the product of its frequency and wavelength, v = νλ.
More free CBSE study material
Notes, worksheets, PPTs and practice papers for Classes 6 to 10 are available at uniquestudyonline.com.
Solutions prepared by Unique Study Point by Sumeet Sahu. Figures are from the NCERT Science textbook Exploration (Class 9).
📋 Details
| Class | Class IX (CBSE / NCERT) |
| Subject | Science |
| Chapter | Chapter 10: Sound Waves: Characteristics and Applications |
| Resource Type | NCERT Solutions |
| Last Updated | 26 October 2026 |
| Session | 2026-27 (Latest NCERT Syllabus) |
| Downloads | 0+ |
| Prepared by | Sumeet Sahu, Unique Study Point, Indore |
| Cost | Free |