NCERT Solutions for Class 9 Science Exploration Chapter 10

Sound is a form of energy produced by vibrating objects and transferred through a material medium as a mechanical wave.
Sound waves travel as compressions and rarefactions, and their properties are described using wavelength, frequency, time period, amplitude, intensity, and speed.

NCERT Solutions for Class 9 Science Exploration Chapter 10 help students solve Sound Waves: Characteristics and Applications from the Class 9 Science textbook. This chapter explains how sound is produced, why it needs a medium, how sound waves travel, how wavelength and frequency are related, why echoes occur, and how ultrasonic waves are used in sonar and echolocation. These NCERT Class 9 Science Solutions include Pause and Ponder answers and all Class 9 Science Chapter 10 exercise solutions in a direct format.

Key Takeaways

• Sound production: Sound is produced by vibrating objects.
• Sound propagation: Sound needs a material medium and cannot travel through vacuum.
• Sound wave type: Sound is a longitudinal mechanical wave.
• Applications: Echo, reverberation, sonar, echolocation, ultrasound, and infrasound are important applications of sound waves.

NCERT Solutions for Class 9 Science Exploration Chapter 10 Structure 2026

NCERT Solutions for Class 9 Science Exploration Chapter 10 In-Text Questions

Class 9 Science Exploration Chapter 10 uses rubber bands, tuning forks, water, slinky, graphs, echoes, sonar, bats, and ultrasonic waves to explain sound waves and their applications.

Q1. Explore various ways of producing sound.

Answer: Sound can be produced by making objects vibrate.

Explanation:
Different objects produce sound in different ways, but vibration is common in all cases.

Examples:

1. Plucking a rubber band produces sound due to vibration.
2. Striking a metal plate produces sound due to vibration.
3. Blowing air into a flute produces sound due to vibration of air column.
4. Speaking produces sound due to vibration of vocal cords.
5. Beating a drum produces sound due to vibration of the stretched membrane.

So, production of sound Class 9 questions are based on the idea that sound is produced by vibrating sources.

Q2. List musical instruments and their vibrating parts.

Answer: Musical instruments produce sound when strings, membranes, air columns, or metal parts vibrate.

Q3. Assertion and Reason: Bell in vacuum jar

Question:
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.

Answer: The correct option is (ii) Both A and R are true, and R is the correct explanation of A.

Explanation:
Sound needs a medium such as solid, liquid, or gas to travel.

When air is pumped out of the jar, there is almost no medium left. The bell can still be seen ringing, but its sound cannot travel to our ears.

This proves that sound cannot travel through vacuum.

Q4. Assertion and Reason: Compressions and rarefactions

Question:
Assertion (A): Compressions and rarefactions move through the medium.
Reason (R): Individual particles of the medium continuously move forward with the wave.

Answer: The correct option is (iii) A is true, but R is false.

Explanation:
In a sound wave, compressions and rarefactions move through the medium.

However, particles of the medium do not move forward with the wave. They only vibrate about their mean positions.

So, the disturbance travels, not the particles.

Q5. What actually reaches your ear when sound travels from a tuning fork?

Question: When sound travels from a tuning fork to your ear, which actually reaches your ear?
(i) Air particles near the tuning fork
(ii) Energy carried by sound waves
(iii) The tuning fork material
(iv) A continuous stream of compressed air

Answer: The correct option is (ii) Energy carried by sound waves.

Explanation:
The air particles near the tuning fork do not travel all the way to the ear. They vibrate and pass the disturbance to neighbouring particles.

The energy of vibration travels through the medium as sound waves.

So, energy carried by sound waves reaches the ear.

Q6. Label compressions and rarefactions in sound wave graphs.

Answer: Regions of higher density are compressions, and regions of lower density are rarefactions.

Explanation:
In a density-distance graph of a sound wave:

1. Crest represents maximum density.
2. Trough represents minimum density.
3. Compression is marked at high-density regions.
4. Rarefaction is marked at low-density regions.

The x-axis should be labelled distance, and the y-axis should be labelled density.

Q7. Compare sounds from thick and thin rubber bands.

Answer: A thin rubber band usually vibrates faster than a thick rubber band and produces a higher-frequency sound.

Explanation:
If the thin rubber band vibrates faster, it completes more oscillations per second.

So:

1. Frequency increases.
2. Time period decreases.
3. Sound is heard at a higher pitch.

A thicker rubber band vibrates more slowly and usually produces a lower-pitch sound.

Q8. How many oscillations does a 20 Hz piston complete per minute?

Answer: The piston completes 1200 oscillations per minute.

Explanation:
Frequency:

$20 , Hz = 20 , oscillations , per , second$

Time:

$1 , minute = 60 , seconds$

Number of oscillations:

$20 \times 60 = 1200$

So, the piston completes 1200 oscillations in one minute.

Q9. Find half of the wavelength from the graph.

Question: For the sound wave represented by the graph, what is half of its wavelength?

Answer: Half of the wavelength is $1.5 , cm$.

Explanation:
From the graph, the distance between consecutive crests or troughs is $3.0 , cm$.

So:

$\lambda = 3.0 , cm$

Half wavelength:

$\frac{\lambda}{2} = \frac{3.0}{2} = 1.5 , cm$

Q10. Compare speed of sound in water, air, steel, and water.

Question: Speed of sound: steel = $5000 , m , s^{-1}$, water = $1500 , m , s^{-1}$, air = $340 , m , s^{-1}$. Find ratios.

Answer:
(i) Speed in water : speed in air = about $4.41 : 1$
(ii) Speed in steel : speed in water = about $3.33 : 1$

Explanation:

For water with respect to air:

$\frac{1500}{340} \approx 4.41$

So, sound travels about 4.41 times faster in water than in air.

For steel with respect to water:

$\frac{5000}{1500} \approx 3.33$

So, sound travels about 3.33 times faster in steel than in water.

Q11. Calculate time difference for sound through air and steel.

Question: Two friends are standing $340 , m$ apart along a steel fence. Calculate the time difference between sound reaching through air and steel.

Answer: The time difference is about $0.932 , s$, so the two sounds can be heard separately.

Explanation:
Distance:

$340 , m$

Time through air:

$t = \frac{distance}{speed}$

$t_{air} = \frac{340}{340} = 1 , s$

Time through steel:

$t_{steel} = \frac{340}{5000} = 0.068 , s$

Time difference:

$1 - 0.068 = 0.932 , s$

Since $0.932 , s > 0.1 , s$, Gunjan can distinguish the two sounds.

Q12. Find minimum distance for echo after 0.2 s.

Question: An experiment requires echoes to arrive at least $0.2 , s$ after the emission of sound. What minimum distance should the reflecting surface be placed at? Speed of sound = $343 , m , s^{-1}$.

Answer: The reflecting surface should be at least $34.3 , m$ away.

Explanation:
Echo time includes travel to the surface and back.

Total distance travelled by sound:

$distance = speed \times time$

$distance = 343 \times 0.2 = 68.6 , m$

Distance of reflecting surface:

$\frac{68.6}{2} = 34.3 , m$

Q13. Find ocean depth using sonar.

Question: A sonar signal takes $4 , s$ to return. Speed of sound in seawater is $1500 , m , s^{-1}$. Find the ocean depth.

Answer: The ocean depth is $3000 , m$.

Explanation:
The sonar signal travels down and returns, so total time is for a round trip.

Time for one-way travel:

$\frac{4}{2} = 2 , s$

Depth:

$distance = speed \times time$

$distance = 1500 \times 2 = 3000 , m$

So, the ocean depth is $3000 , m$.

NCERT Solutions for Class 9 Science Exploration Chapter 10 Exercise Questions

The Revise, Reflect, Refine section includes MCQs, graph-based questions, echo and reverberation, wavelength, frequency, time period, sonar, and speed of sound numericals.

Q1. Which observation supports that sound is a mechanical wave?

Answer: The correct option is (ii) Sound needs a medium to propagate.

Explanation:
Mechanical waves require a material medium for propagation. Sound travels through solids, liquids, and gases, but it cannot travel through vacuum.

So, the need for a medium best supports that sound is a mechanical wave.

Q2. What increases when frequency of a sound wave increases?

Answer: The correct option is (iii) number of compressions per second.

Explanation:
Frequency means the number of oscillations or compressions passing a point per second.

If frequency increases, more compressions pass a point each second.

For a fixed medium, speed remains nearly constant, so wavelength decreases when frequency increases.

Q3. Find frequency if 20 compressions pass in 4 seconds.

Answer: The correct option is (ii) $5 , Hz$.

Explanation:
Frequency:

$\nu = \frac{Number , of , oscillations}{Time}$

$\nu = \frac{20}{4}$

$\nu = 5 , Hz$

Q4. Reflected sound reaches after 0.05 s. Echo or reverberation?

Answer: It will produce reverberation, not a clear echo.

Explanation:
A clear echo is heard when the reflected sound reaches the ear at least $0.1 , s$ after the original sound.

Here, the reflected sound reaches after $0.05 , s$, which is less than $0.1 , s$.

So, the reflected sound mixes with the original sound and causes reverberation.

Q5. Compare wavelength and amplitude from two graphs.

Answer: The wave with more distance between two consecutive crests has greater wavelength. The wave with smaller height from average density has smaller amplitude.

Explanation:
If the scales on both axes are the same:

1. Greater wavelength is shown by wider spacing between crests or troughs.
2. Smaller amplitude is shown by a lower maximum change in density from the average line.

So, identify wavelength from horizontal spacing and amplitude from vertical height.

Q6. Identify curves A, B, and C based on frequency.

Answer: The curve with the shortest wavelength is A, the curve with medium wavelength is B, and the curve with the longest wavelength is C.

Explanation:
For waves travelling with the same speed:

$v = \lambda \nu$

Frequency is inversely proportional to wavelength.

So:

1. Maximum frequency means shortest wavelength.
2. Minimum frequency means longest wavelength.
3. Intermediate frequency means intermediate wavelength.

Q7. Draw graph for amplitude 3 units and wavelength 4 cm.

Answer: Draw a wave whose maximum height above the average line is 3 units and distance between two consecutive crests is 4 cm.

Explanation:
Use these labels:

1. x-axis: Distance in cm
2. y-axis: Density variation
3. Average density line: centre horizontal line
4. Amplitude: 3 units above and below average line
5. Wavelength: 4 cm between consecutive crests

The graph should show repeating crests and troughs with equal spacing.

Q8. What is wrong in showing sound with a spacecraft explosion in space?

Answer: The error is that sound cannot travel through space because space is almost vacuum.

Explanation:
A flash of light from the explosion can be seen because light does not need a medium.

Sound cannot be heard directly in space because sound needs a material medium to propagate.

So, showing the explosion sound reaching the observer at the same time as light is scientifically incorrect.

Q9. Find time period when wavelength is 3.44 m and speed is $344 , m , s^{-1}$.

Answer: The time period is $0.01 , s$.

Explanation:
Use:

$v = \lambda \nu$

$\nu = \frac{v}{\lambda}$

$\nu = \frac{344}{3.44} = 100 , Hz$

Time period:

$T = \frac{1}{\nu}$

$T = \frac{1}{100} = 0.01 , s$

Q10. Find depth of sunken ship using sonar.

Question: A sonar echo is detected after $5 , s$. Speed of ultrasonic wave in seawater is $1525 , m , s^{-1}$. Find depth.

Answer: The wreckage is approximately $3812.5 , m$ deep.

Explanation:
Total time is for downward and upward travel.

One-way time:

$\frac{5}{2} = 2.5 , s$

Depth:

$distance = speed \times time$

$distance = 1525 \times 2.5$

$distance = 3812.5 , m$

Q11. Find time taken by ultrasonic parking sensor.

Question: A vehicle sensor starts warning at $1.2 , m$ from an obstacle. Speed of ultrasonic wave in air is $345 , m , s^{-1}$. Find time for the wave to go and return.

Answer: The ultrasonic wave takes about $0.00696 , s$.

Explanation:
Distance to obstacle and back:

$2 \times 1.2 = 2.4 , m$

Time:

$t = \frac{distance}{speed}$

$t = \frac{2.4}{345}$

$t \approx 0.00696 , s$

So, the time is about $6.96 \times 10^{-3} , s$.

Q12. Find extra time for thunder at $0^\circ C$ compared with $22^\circ C$.

Question: Speed of sound is $331 , m , s^{-1}$ at $0^\circ C$ and $344 , m , s^{-1}$ at $22^\circ C$. Find extra time for thunder to travel $1720 , m$ at $0^\circ C$.

Answer: The extra time is about $0.20 , s$.

Explanation:
Time at $22^\circ C$:

$t_1 = \frac{1720}{344} = 5.00 , s$

Time at $0^\circ C$:

$t_2 = \frac{1720}{331} \approx 5.20 , s$

Extra time:

$5.20 - 5.00 = 0.20 , s$

Q13. Calculate wavelength and frequency from Fig. 10.32.

Question: A sound wave travels with speed $340 , m , s^{-1}$. The graph shows a wavelength of $8 , cm$. Find wavelength and frequency.

Answer: The wavelength is $0.08 , m$ and the frequency is $4250 , Hz$.

Explanation:
Convert wavelength:

$8 , cm = 0.08 , m$

Use:

$v = \lambda \nu$

$\nu = \frac{v}{\lambda}$

$\nu = \frac{340}{0.08}$

$\nu = 4250 , Hz$

Q14. Find wavelengths and frequencies of waves A and B.

Question: Two sound waves A and B travel at the same speed, $345 , m , s^{-1}$. Use Fig. 10.33.

Answer: Wave A has wavelength $5.0 , cm$ and frequency $6900 , Hz$. Wave B has wavelength $2.5 , cm$ and frequency $13800 , Hz$.

Explanation:
From the graph:

$\lambda_A = 5.0 , cm = 0.05 , m$

$\lambda_B = 2.5 , cm = 0.025 , m$

For wave A:

$\nu_A = \frac{v}{\lambda_A}$

$\nu_A = \frac{345}{0.05} = 6900 , Hz$

For wave B:

$\nu_B = \frac{345}{0.025} = 13800 , Hz$

Wave B has the shorter wavelength and higher frequency.

Q15. Find ratio of speed of sound in air and water.

Question: Two identical sound sources are placed at A in air and B in water. Time taken by sound to return to A is 4.5 times that of B. Find ratio of speeds in air and water.

Answer: The ratio of speed of sound in air to water is $1 : 4.5$.

Explanation:
Both waves travel the same distance to the cliff and back.

For the same distance:

$Speed \propto \frac{1}{Time}$

Given:

$t_A = 4.5t_B$

So:

$\frac{v_A}{v_B} = \frac{t_B}{t_A}$

$\frac{v_A}{v_B} = \frac{1}{4.5}$

Thus:

$v_{air} : v_{water} = 1 : 4.5$

NCERT Solutions for Class 9 Science Exploration

Topics Covered in NCERT Solutions for Class 9 Science Exploration Chapter 10

Class 9 Science Exploration Chapter 10 covers sound production, sound propagation, sound wave characteristics, reflection of sound, and applications of ultrasonic and infrasonic waves.

• Production of sound Class 9
• Vibrating objects as sources of sound
• Vocal cords and sound production in humans
• Tuning fork and vibration
• Propagation of sound Class 9
• Sound through solids, liquids, and gases
• Sound needs a medium
• Vacuum bell jar experiment
• Sound waves Class 9 Science
• Compressions and rarefactions
• Longitudinal waves and mechanical waves
• Energy of sound waves
• Graphical representation of sound waves
• Wavelength frequency and time period Class 9
• Amplitude and intensity of sound Class 9
• Speed of sound Class 9
• Pitch and loudness
• Human audible range
• Echo and reverberation Class 9
• Ultrasonic and infrasonic waves Class 9
• Sonar and echolocation Class 9

Important Concepts in NCERT Solutions for Class 9 Science Exploration Chapter 10

Sound Waves Characteristics and Applications Class 9 requires students to connect sound production with vibration, medium, wave motion, graphs, and real-life uses.

Important Formulas in NCERT Solutions for Class 9 Science Exploration Chapter 10

Class 9 Science Chapter 10 solutions use formulas for frequency, time period, speed of sound, echo, and sonar calculations.