Physics

NCERT QUESTIONS WITH SOLUTIONS

1. How does the sound produced by a vibrating This region is called rarefaction.
object in a medium reach your ear? Collisions among the air particles cause
Solution the pressure variations to move away
Air is the most common material through from the bell in all directions. If you were
which sound propagates. When a to focus at one spot, you would see the
vibrating object like prongs of tuning fork value of the air pressure rise and fall. In
move forward, they push the molecules of this way, the pressure variations i.e., the
the air in front of them. This is turn sound is produced by the bell which is
compresses the air, thus creating a region transmitted through air (see figure).
of high pressure and high density called
compression. This compression in the air
travels forward. When the prongs of the
(a) Before the bell is struck, the air around
tuning fork move backward, they create a it is a region of average pressure
region of low pressure and low density in
the air, commonly called rarefaction. As
the tuning fork continues to vibrates, it
Vibrating bell R C R C R C
produces a series of successive
(b) Once the bell is struck, the vibrating edge
compressions and rarefactions in the air, creates regions of high and low pressure
thus, propagating sound through the air 3. Why are sound waves called mechanical
which finally reaches our ears.
waves?
2. Explain how sound is produced by your
Solution
school bell.
The waves which require a medium for
Solution
their propagation are called mechanical
When we hit a school bell by a hard gong,
waves. Sound waves also propagate
it moves back and forth about its
through a medium because of the
equilibrium position. Thus, the edge of
interaction of the particles present in that
the bell strikes the particles in the air.
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medium. Sound waves force the medium

When the edge moves forward, air
particles to vibrate. Hence, these waves
particles are driven forward. This
are known as mechanical waves.
forward motion of the bell produces a
4. Suppose you and your friend are on the
region where the air pressure is slightly
moon. Will you be able to hear any sound
higher than average. This region is called
produced by your friend?
compression. When the edge moves
backward, air particles near it have more Solution
space and thus, they spread farther apart. No, I will not be able to hear sound,
This backward motion produces a region because Moon has no atmosphere.
where the air pressure is slightly below Therefore, no sound waves can travel to
average pressure. my ears and therefore, no sound is heard.

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NCERT Basics : Class 9
5. Which wave property determines 9. Calculate the wavelength of a sound wave
(a) loudness (b) pitch? whose frequency is 220 Hz and speed is
Solution 440 ms–1 in a given medium.
(a) The amplitude of the wave Solution
determines the loudness; more the Given, frequency,  = 220 Hz ;
amplitude of a wave, more is the speed of sound, v = 440 ms–1
loudness produced. v 440
Speed, v =   or  = = =2m
(b) The pitch is determined by the  220
frequency of the wave. Higher the 10. A person is listening to a tone of 500 Hz
frequency of a wave, more sitting at a distance of 450 m from the
is its pitch and shriller is the sound. source of the sound. What is the time
6. Guess which sound has a higher pitch: interval between successive
guitar or car horn? compressions from the source ?
Solution Solution
The frequency of vibration of a sound The time interval between two successive
produced by a guitar is greater than that compressions is equal to the time period
produced by a car horn. Since the pitch of of the wave. It has nothing to do with the
a sound is proportional to its frequency, distance (450 m) of the person from the
the guitar has a higher pitch than a car source.
horn. 1 1
Time interval = =
7. What are wavelength, frequency, time Frequency 500
period and amplitude of a sound wave? = 2 × 10–3 s
Solution 11. Distinguish between loudness and
Wavelength : It is the distance between intensity of sound.
two consecutive compressions or two Solution
consecutive rarefactions of a wave. Intensity is an objective property of the
Frequency : The number of sound wave. In fact, it is related to the
compressions or rarefactions taken square of the wave amplitude, and does
together passing through a point in one not depend on the particular
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second is called frequency. characteristics of a person’s ears. The

Time Period : It is the time taken by two amount of sound energy passing each
consecutive compressions or rarefactions second through unit area is called the
to cross a point. intensity of sound. Loudness, on the other
Amplitude : It is the magnitude of hand, is a subjective property of the
maximum displacement of a vibrating sound that depends on the human ear, the
particle about its mean position. sensitivity of the ear to the frequency of
8. How are the wavelength and frequency of the sound, and the distance from the
a sound wave related to its speed? source of the sound. In other words,
loudness can be considered as the
Solution
intensity of an audible sound.
Speed of sound = frequency × wavelength
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 Physics
12. In which of the three media, air, water or Solution
iron; does sound travel the fastest at a (a) Infrasound : Sound waves of
particular temperature? frequencies below 20 Hz.
Solution (b) Ultrasound : Sound waves of
Sound travels the fastest in solids. Its frequencies above 20,000 Hz.
17. A submarine emits a sonar pulse, which
speed decreases in liquids and it is the
returns from an underwater cliff in 1.02 s.
slowest in gases. Therefore, for a given
If the speed of sound in salt water is
temperature, sound travels fastest in
1531 ms–1, how far away is the cliff ?
iron.
Solution
13. An echo is returned in 3 s. What is the Given, time taken by the sonar pulse,
distance of the reflecting surface from the t = 1.02 s ; speed of sound in salt water,
source, given that the speed of sound is v = 1531 m s–1 ;
342 ms–1 ? distance of cliff, s = ?
Solution v× t
Now, s =
Given, speed of sound, v = 342 ms–1 ; time 2
taken for hearing the echo, T = 3 S ; 1531 × 1.02
=
2
distance of the reflecting surface, s = ?
= 780.81 m
v × t 342 × 3
Now, s = = 18. What is sound and how is it produced ?
2 2
Solution
= 513 m
Sound is a form of energy which produces
14. Why is the ceiling of concert halls curved ?
sensation of hearing. When an object is
Solution set into vibrations, sound is produced.
The ceiling of concert halls is curved so 19. Describe with the help of a diagram, how
that sound after reflection from it reaches compressions and rarefactions are
all the corners of the hall and is audible to produced in air near a source of sound.
every person in the hall. Solution
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15. What is the audible range of the average When a vibrating body like a tuning fork
human ear ? moves forward, it creates a region of high
Solution pressure in its vicinity. This region of high
An average human ear can hear sound pressure is known as compression. When
it moves backward, it creates a region of
waves of frequencies between 20 Hz to
low pressure in its vicinity. This region is
20,000 Hz.
known as a rarefaction. As the body
16. What is the range of frequencies
continues to move forward and
associated with
backward, it produces a series of
(a) Infrasound ? compressions and rarefactions
(b) Ultrasound ? (see figure).

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NCERT Basics : Class 9
Solution
The speed of light is 3 × 108 ms–1 and the
speed of sound is 344 ms–1 in air. That is,
(a) No vibrations in the tuning forks the speed of light is very large as
compared to the speed of sound. Thus,
flash of lightning is seen at first, but sound
takes a little bit more time to reach our
Compression (c)
ears.
(b) Forward movement in the tuning fork
24. A person has a hearing range from 20 Hz
to 20 kHz. What are the typical
wavelengths of sound waves in air
corresponding to these two frequencies?
Rarefaction (R)
Take the speed of sound in air as
(c) Backward movement in the tuning fork
344 ms–1.
Solution
Given, speed of sound, v = 344 ms–1 , For
frequency 1 = 20 Hz, wavelength, 1 = ?
R C R C R C
Wavelength of sound of frequency 20 Hz
(d) Vibrations in tuning fork produces v 344
Series of compressions and rarefactions 1 = = = 17.2 m
1 20
21. Why is sound wave called a longitudinal
wave? For frequency 2 = 20 kHz = 20000 Hz,
wavelength, 2 = ?
Solution
v 344
Sound wave is called longitudinal wave 2 = = = 1.72 × 10–2 m
2 20000
because the particles of the medium
25. Two children are at opposite ends of an
vibrate in the direction of the propagation
aluminium rod. One strikes the end of the
of wave.
rod with a stone. Find the ratio of times Publishing\PNCF\2024-25\LIVE Module\SET-2\NCERT\9th\Physics\6. Sound

22. Which characteristic of the sound helps taken by the sound wave in air and in
you to identify your friend by his voice aluminium to reach the second child.
while sitting with others in a dark room? Given, velocity of sound in air and
Solution aluminium are 346 ms–1 and 6420 ms–1
The characteristic of sound that helps you respectively.
Solution
to identify your friend's voice is 'quality'
Here, the distance travelled by sound
or 'timbre'.
waves through air and aluminium is same
23. Flash and thunder are produced
which is equal to the length (s) of the rod.
simultaneously. But thunder is heard a For air, let the speed be v1 and time be t1.,
few seconds after the flash is seen. Why? For Al, let the speed be v2 and time be t2.
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 Physics
s in our brain for about 0.1 s). Thus, no echo
Now, v 1 = or s = v1t1 ----- (1)
t1 is heard.
s 29. Give two practical applications of
Also, v 2 = or s = v2t2 ----- (2) reflection of sound waves.
t2
Solution
From (1) & (2), we get, v1t1 = v2t2
(i) Megaphones which are designed to
t 1 v2 6420 send sound waves in particular
= = = 18.55
t 2 v1 346 direction are based on the reflection
26. The frequency of a source of sound is 100 of sound.
Hz. How many times does it vibrate in a (ii) In stethoscope the sound of patient’s
minute ? heartbeat reaches the doctor’s ears
Solution by multiple reflection in the tubes.
No. of vibrations produced in 1 s = 100 30. A stone is dropped from the top of a tower
No. of vibrations produced in 1 minute 500 m high into a pond of water at the
i.e., 60 seconds = 100 × 60 = 6000. base of the tower. When is the splash
Thus, the source of sound vibrates 6000 heard at the top ? Given, g = 10 ms–2 ;
times in a minute. speed of sound = 340 m/s.
27. Does sound follow the same laws of Solution
reflection as light does ? Explain. For the downward journey of stone,
Solution Initial velocity, u = 0,
Distance travelled, s = height of tower =
Yes, sound and light follow the same laws
500 m, Time of fall, t1 =?, g = 10 ms–2
of reflection given below :
1 1
(a) Angle of incidence = Angle of s = ut + gt2 or 500 = 0 × t1 + ×
2 2
reflection.
10 × t12
(b) The incident sound wave, the
or 500 = 5t12 or t12 = 100
normal and the reflected sound
t1 = 10 s
wave lie in the same plane. For the sound travelling upward, Time
28. When a sound is reflected from a distant taken, t2 = ?
object, an echo is produced. Let the There is no effect of gravity on the
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distance between the reflecting surface propagation of sound i.e, always the
and the source of sound production formula of uniform motion is used for
remain the same. Do you hear echo sound sound or any other wave.
on a hotter day? s
Speed of sound, v=
Solution t2
If the temperature rises, the speed of s 500
sound will increase. Thus, the time taken or t 2 = = = 1.47 s
v 340
by the sound to reach initial point  Total time = t1 + t2 = 10 + 1.47
decreases. That is, it will be certainly less = 11.47 s
than 0.1 s (the sensation of sound persists 31. A sound wave travels at a speed of 339
ms–1. If its wavelength is 1.5 cm, what is

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NCERT Basics : Class 9
the frequency of the wave? Will it be Loudness of the sound depends inversely
audible ? on the square of the distance of the
Solution observer from the source of sound, and is
Speed of sound wave, v = 339 ms–1 ;
directly proportional to the square of the
Wavelength,  = 1.5 cm = 0.015 m ;
amplitude of the audible sound wave.
Frequency,  = ?
34. Explain how bats use ultrasound to catch
v 339
Frequency,  = = a prey.
 0.015
= 22600 Hz Solution
The sound will not be audible, because The bats produce high pitched ultrasonic
human beings can hear only up to 20,000 waves which are not heard by human
Hz. beings. The ultrasonic waves on striking
32. What is reverberation? How can it be the insect send back an echo, which is
reduced? heard by the bat. As the echo is heard by
Solution the bat, it moves towards the insect and
Reverberation is the repeated/multiple catches it.
reflections of sound in any big enclosed 35. How is ultrasound used for cleaning ?
space. It can be reduced by covering the Solution
ceiling and walls of the enclosed space The object to be cleaned is put in a tank
with some sound-absorbing materials
fitted with ultrasonic vibrator. The tank is
like fibre board, loose woolens etc.
filled with water containing detergent. As
33. What is loudness of sound ? What factors
does it depend on? the ultrasonic vibrator is switched on, the
Solution detergent rubs against the object at a very
The loudness of the sounds humans high speed and hence cleans it.
perceive relates to the intensity of the 38. Explain how defects in a metal block can
audible sound. Loudness is a subjective be detected using ultrasound.
property of the sound that depends on Solution
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the human ear, the sensitivity of the ear Refer theory, subtopic 14.
to the frequency of the sound, and the
distance from the source of the sound.

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