NCERT Exemplar Class 9 Science Solutions Chapter 12 Sound 2021

NCERT Exemplar Class 9 Science Solutions Chapter 12 Sound

Edited By Vishal kumar | Updated on Apr 15, 2025 12:48 PM IST

Have you wondered how fast sound travels, how we measure the loudness of sound and what makes a melody sound good? Explore the fun side of science with Class 9 NCERT Sound Solutions – where you learn how sound travels in a super easy way! The chapter will teach you about the important parameters required to study sound such as loudness and pitch, and how they relate to sound waves.

This Story also Contains

NCERT Exemplar Class 9 Science Solutions Chapter 12 - MCQ
NCERT Exemplar Class 9 Science Solutions Chapter 12 - Short Answer
NCERT Exemplar Class 9 Science Solutions Chapter 12-Long Answer
NCERT Exemplar Class 9 Science Solutions Chapter 12 Important Topics:
NCERT Class 9 Science Exemplar Solutions for Other Chapters:
Features of NCERT Exemplar Class 9 Science Solutions Chapter 12:

NCERT Exemplar Class 9 Science Solutions Chapter 12 Sound

These expert answers help you get ready for competitive examinations while making tough concepts simple to understand. The NCERT Exemplar Solutions for Class 9 Science Chapter 12 – Sound lets students understand how sound is being produced and the various factors involved. These solutions are specially made by experts at Careers360 for students who are preparing for competitive exams like JEE Main or NEET. These solutions make the concepts easy to learn and grasp for students.

NCERT Exemplar Class 9 Science Solutions Chapter 12 - MCQ

Question:1

Note is a sound
(a) of mixture of several frequencies
(b) of mixture of two frequencies only
(c) of a single frequency
(d) Always unpleasant to listen

Answer: (a)
Solution:
Sound of a single frequency is called tone.
Sound of mixture of multiple frequencies is called note.
This mixture may have more than two frequencies.
Frequency of sound is also known as pitch.
Frequency depends on the structure of a sound producing device.
That is why, it is said that frequency depends on the source of the sound.
The correct answer of this question is A.

Question:2

A key of a mechanical piano struck gently and then struck again but much harder this time. In the second case
(a) sound will be louder but pitch will not be different
(b) sound will be louder and pitch will also be higher
(c) sound will be louder but pitch will be lower
(d) both loudness and pitch will remain unaffected

Answer: (A)
Solution:
Frequency of sound depends on the mechanical structure of sound producing device.
Different keys of piano will produced sound of different frequency. A single key will always produce sound of same frequency. If we strike the key with force, the frequency will not change but amplitude of sound will change and it will lead to change in loudness.
Frequency is also known as pitch, so we can say if we struck harder the sound will be louder but pitch will be same.
Hence the correct option is option A

Question:3

In SONAR, we use
(a) ultrasonic waves
(b) infrasonic waves
(c) radio waves
(d) audible sound waves

Answer: A
Solution:
Sonar is the navigation technique used to navigate the location of submarine underwater and it uses Ultrasonic waves.

Question:4

Sound travels in air if
(a) Particles of medium travel from one place to another
(b) There is no moisture in the atmosphere
(c) Disturbance moves
(d) Both particles as well as disturbance travel from one place to another.

Answer:

Answer: C
Solution:
When sound travels in a medium, molecules of medium does not go from one place to another.
The disturbance due to sound is transferred from one molecule to another, hence we can say disturbance travels when sound travels.
In presence of moisture, the speed of sound in air increases. Therefore, sound can certainly travel in moist atmosphere.
Hence the correct answer of this question is option C

Question:5

When we change feeble sound to loud sound we increase its
(a) frequency
(b) amplitude
(c) velocity
(d) wavelength

Answer:

Answer: B
Solution:
Generally sound is characterized by its pitch and loudness.
Pitch is frequency of sound, which tells us number of oscillations, any particle will make in one second.

f = \frac{1}{T}

The loudness represent energy or intensity of the sound.
It is generally governed by the fact that how far a particle moves in its oscillation.
This maximum displacement of the particle in oscillation is called amplitude.
So loudness increases means amplitude of oscillation increases.
Hence the correct answer of this question is option B

Question:6

In the curve (Fig.12.1) half the wavelength is
(a) A B
(b) B D
(c) D E
(d) A E

Answer:

Answer: B
Solution:
Distance between two consecutive crests or consecutive troughs is equal to wavelength of any sound.
In the given figure, the separation between A and E will be equal to one wavelength.
Therefore, half wave length will be equal to AC or CE or BD distance.
Hence, the correct answer of this question is option B

Question:7

Earthquake produces which kind of sound before the main shock wave begins
(a) ultrasound
(b) infrasound
(c) audible sound
(d) none of the above

Answer:

Answer: B
Solution:
Earthquake produces Seismic waves before the main shock wave.
The Seismic waves have frequency less than 10 Hz.
This frequency is less than audible range and it is called infra sound or infrasonic wave.
The sound of frequency more than that of audible range is called ultrasound or ultrasonic waves.
Sound of frequency less than 20 Hz is called infrasonic and a frequency more than 20,000 Hz is called ultrasonic.
The correct answer of this question is option B

Question:8

Infra sound can be heard by
(a) dog
(b) bat
(c) rhinoceros
(d) human beings

Answer:

Answer: C
Solution:
The frequency range, which human being can hear is called audible sound.
Sound of frequency less than 20 Hz is called infrasonic and a frequency more than 20,000 Hz is called ultrasonic.
Rhinoceros can sense sound of frequency less than audible range which is called infra sound.
The correct answer of this question is option C.

Question:9

Before playing the orchestra in a musical concert, a sitarist tries to adjust the tension and pluck the string suitably. By doing so, he is adjusting:
(a) intensity of sound only
(b) amplitude of sound only
(c) frequency of the sitar string with the frequency of other musical instruments
(d) loudness of sound

Answer:

Answer: C
Solution:
The frequency of sound generated from the string, depends on the length of the string, the tension in the string as well as thickness of the string.
Hence, by adjusting the strings of sitar, the Sitarist adjusts frequency with other musical instruments.
Loudness, intensity or amplitude all represent the same physical phenomenon and it depends on how hard you strike the string.
Hence the correct option is option C

NCERT Exemplar Class 9 Science Solutions Chapter 12 - Short Answer

Question:10

The given graph (Fig.12.2) shows the displacement versus time relation for a disturbance travelling with velocity of 1500 m s^–1. Calculate the wavelength of the disturbance.
q10

Answer:

In propagation of sound, the particle of medium oscillates.
By the graph, we can see that a time period of one complete oscillation is to 2 microseconds.
The wavelength of disturbance is defined as: distance travelled by disturbance in one time period of oscillation.

λ = V \times T

λ = \frac{V}{f}

As the disturbance is moving with 1500 m/s and time period is 2 microseconds, The wavelength can be calculated as:

λ = V \times T = 1500 \times 2 \times 10^{- 6} = 3 \times 10^{- 3} = 3 m m

Question:11

Which of the above two graphs (a) and (b) (Fig.12.3) representing the human voice is likely to be the male voice? Give reason for your answer.
q11

Answer:

The difference between human male voice and female voice is on the basis of pitch or frequency.
Male voice will have lower frequency in comparison with female voice.
We know that frequency is inverse of time period of oscillation.

f = \frac{1}{T}

In the given graph, it is quite visible that time period of graph (a) is more than time period of graph (b).
Therefore graph (a) must be corresponding to male voice.

Question:12

A girl is sitting in the middle of a park of dimension 12 m × 12 m. On the left side of it there is a building adjoining the park and on right side of the park, there is a road adjoining the park. A sound is produced on the road by a cracker. Is it possible for the girl to hear the echo of this sound? Explain your answer.

Answer:

When we observe a sound directly or after reflection from any building, we call it an echo.
This echo can be heard or these two sounds can be distinguished, only if time gap between them is more than 0.1 second.
For this, separation between source and the wall must be more than 16 meter.
Assuming speed of sound = 320m/s
Distance between source and wall is d, then time to reach reflected sound will be:

t = \frac{2 d}{V}

In the given case, the size of park is small. Therefore, it cannot cause any echo of the sound produced on the road.

Question:13

Why do we hear the sound produced by the humming bees while the sound of vibrations of pendulum is not heard?

Answer:

The human ear can sense sound only if its frequency is more than 20 Hz and less than 20,000 Hz.
Sound of frequency less than 20 Hz is called infrasonic and a frequency more than 20,000 Hz is called ultrasonic.
The frequency of the disturbance created by humming bee is more than 20 Hz that’s why we can hear the sound.
Due to the vibration of pendulum, the frequency of disturbance is less than 20 Hz and we cannot listen the sound of vibration of pendulum.

Question:14

If any explosion takes place at the bottom of a lake. Infra sound type of shock waves in water will take place?

Answer:

Transverse wave cannot travel in air or within the water.
Transverse wave can only travel at the surface of water or in solids.
So when any explosion takes place at the bottom of lake, the shock waves have to travel within the water.
These shock waves cannot be transverse and they have to be longitudinal waves like sound wave.

Question:15

Sound produced by a thunderstorm is heard 10 s after the lightning is seen. Calculate the approximate distance of the thunder cloud. (Given speed of sound = 340 m s^–1.)

Answer:

The speed of light is very large in comparison with the speed of sound.
We can assume that lightning can be seen, without any time delay as light travels very fast.
If the sound produced by thunderstorm is heard 10 seconds after lightning that means sound took 10 seconds to reach from thundercloud.
As the speed of sound is 340 m/s, so distance travelled by sound in 10 seconds will be 3400 meters.
So the approximate distance of thunder cloud is 3.4 km.

Question:16

For hearing the loudest ticking sound heard by the ear, find the angle x in the Fig.12.4.
q16

Answer:

Similar to light, sound also follows law of reflection.
If I sound is coming at an incident angle of

∠ i

then it will go at the same reflection angle

∠ i = ∠ r

.
In the given question, sound coming through the first pipe has angle of incidence equal to 40° that can be calculated as (90° - 50°)
The same sound will be reflected at an angle of 40°. Therefore, the second pipe has to be kept at the angle of 40° from normal.
Therefore the value of x will be 40°.

Question:17

Why is the ceiling and wall behind the stage of good conference halls or concert halls made curved?

Answer:

The design of ceiling and wall in a good conference hall is to ensure proper sound at every seat of the conference hall.
This design is called acoustic design and design engineers are called acoustic engineers.
The ceiling and wall behind the stage are made curved, so that the reflected sound from them can reach to the audience evenly.

NCERT Exemplar Class 9 Science Solutions Chapter 12-Long Answer

Question:18

Represent graphically by two separate diagrams in each case
(a) Two sound waves having the same amplitude but different frequencies?
(b) Two sound waves having the same frequency but different amplitudes.
(c) Two sound waves having different amplitudes and also different wavelengths.

Answer:

In propagation of sound, the particle of medium oscillates.
The time to complete one oscillation is called time period T.
Pitch is frequency of sound, which tells us number of oscillations, any particle will make in one second.

f = \frac{1}{T}

Distance between two consecutive crests or consecutive troughs is equal to wavelength of any sound.
By these information, we can draw the corresponding graphs.

1662033469117

Question:19

Establish the relationship between speed of sound, its wavelength and frequency. If velocity of sound in air is 340 m s^–1, calculate
(i) Wavelength when frequency is 256 Hz.
(ii) Frequency when wavelength is 0.85 m.

Answer:

The wavelength of any sound is defined as distance travelled by disturbance in one time period of the oscillation of particle.

λ = V \times T

The number of oscillations per unit time is called frequency. Hence, frequency and time period are related as.

f = \frac{1}{T}

By putting the value of time period in the relation of wavelength, we will get:

λ = \frac{V}{f} \Rightarrow V = f \times λ

i) If frequency and sound velocity is given, wavelength can be calculated as:

λ = \frac{V}{f} \Rightarrow λ = \frac{340}{256} = 1.328125 m

ii) If wavelength and sound velocity is given, frequency of sound can be calculated as:

f = \frac{V}{λ} \Rightarrow f = \frac{340}{0.85} = 400 H z

Question:20

Draw a curve showing density or pressure variations with respect to distance for a disturbance produced by sound. Mark the position of compression and rarefaction on this curve. Also define wavelengths and time period using this curve.

Answer:

In propagation of sound, the particle of medium oscillate.
The time to complete one oscillation is called time period T.
Pitch is frequency of sound, which tells us number of oscillations, any particle will make in one second.

f = \frac{1}{T}

Distance between two consecutive crest or consecutive troughs is equal to wavelength of any sound.
a201

NCERT Exemplar Class 9 Science Solutions Chapter 12 Important Topics:

Topics covered in the chapter via Class 9 Science NCERT Exemplar Chapter 12 solution are:

The student will learn about parameters of a sound wave such as wavelength and frequency.
Students will learn that sound cannot be propagated in vacuum and travels faster in water than in air.
NCERT Exemplar Class 9 Science Solutions Chapter 12 discusses the representation of sound in form of graphs is taught in this chapter.
Students will learn about “crest and trough” and similarly learn about compression and rarefaction zone.

NCERT Class 9 Exemplar Solutions for Other Subjects:

NCERT Class 9 Science Exemplar Solutions for Other Chapters:

Chapter wise solutions

Chapter 1 Matter in our Surroundings

Chapter 2 Is Matter Around Us Pure?

Chapter 3 Atoms and Molecules

Chapter 4 Structure of the Atom

Chapter 5 The Fundamental Unit of Life

Chapter 6 Tissues

Chapter 7 Diversity in Living Organisms

Chapter 8 Motion

Chapter 9 Forces and Laws of Motion

Chapter 10 Gravitation

Chapter 11 Work and Energy

Chapter 13 Why do We Fall ill?

Chapter 14 Natural Resources

Chapter 15 Improvement in Food Resources

Features of NCERT Exemplar Class 9 Science Solutions Chapter 12:

These Class 9 Science NCERT Exemplar Chapter 12 solutions explain the basics of sound and how it travels.
They are very useful for higher classes and exams like JEE and NEET.
You can use these solutions to practice and understand the topic clearly.
They are enough to help you prepare for other books too, like the NCERT textbook or S. Chand.

Check NCERT Solutions for questions given in the book

Check the Solutions of Questions Given in the Book

Chapter No.	Chapter Name
Chapter 1	Matter in Our Surroundings
Chapter 2	Is Matter Around Us Pure
Chapter 3	Atoms and Molecules
Chapter 4	Structure of The Atom
Chapter 5	The Fundamental Unit of Life
Chapter 6	Tissues
Chapter 7	Diversity in Living Organisms
Chapter 8	Motion
Chapter 9	Force and Laws of Motion
Chapter 10	Gravitation
Chapter 11	Work and Energy
Chapter 12	Sound
Chapter 13	Why Do We Fall ill?
Chapter 14	Natural Resources
Chapter 15	Improvement in Food Resources

Also, Read NCERT Solution Subject Wise

Check NCERT Notes Subject Wise

Also, check the NCERT Books and NCERT Syllabus here

Frequently Asked Questions (FAQs)

1. “Frequency is identification of any sound”. comment on the statement.

When sound travels and changes medium, the speed of sound changes. It causes a change in wavelength but not the frequency. Frequency does not depend on the medium.

2. If a person will shout, does that mean the frequency of sound is increased?

No, the frequency of sound produced by any woman does not change on changing the loudness. The intensity of sound increases on shouting but the frequency of sound will remain the same. Frequency is generally known as pitch which can be seen as identification of someone’s voice

3. Can we talk on the moon?

No, sound needs medium to travel and there is no atmosphere on the moon. Therefore, we cannot talk on the moon as we talk on earth.

4. Can I revise using these questions?

Yes! These are perfect for quick and smart revision before your tests.

5. Are these exemplar questions important?

Yes! They help you understand the chapter better and are great for exam practice.

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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NCERT Exemplar Class 9 Science Solutions Chapter 12 Sound

NCERT Exemplar Class 9 Science Solutions Chapter 12 - MCQ

NCERT Exemplar Class 9 Science Solutions Chapter 12 - Short Answer

NCERT Exemplar Class 9 Science Solutions Chapter 12-Long Answer

NCERT Exemplar Class 9 Science Solutions Chapter 12 Important Topics:

NCERT Class 9 Exemplar Solutions for Other Subjects:

NCERT Class 9 Science Exemplar Solutions for Other Chapters:

Features of NCERT Exemplar Class 9 Science Solutions Chapter 12:

Check the Solutions of Questions Given in the Book

Also, Read NCERT Solution Subject Wise

Check NCERT Notes Subject Wise

Also, check the NCERT Books and NCERT Syllabus here

Frequently Asked Questions (FAQs)

Also Read

Articles

Upcoming School Exams

Assam High School Leaving Certificate Examination

National Institute of Open Schooling 10th examination