NCERT Exemplar Class 9 Science Chapter 12 Sound

NCERT Exemplar Class 9 Science Chapter 12 Sound are part of NCERT Exemplar Class 9 Science. Here we have given NCERT Exemplar Class 9 Science Solutions Chapter 12 Sound.

NCERT Exemplar Class 9 Science Solutions Chapter 12 Sound

Multiple Choice Questions

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
Solution:
(a) Note is a sound of mixture of several frequencies.

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
Solution:
(a) When the key of a mechanical piano is struck harder, sound will be louder as it will vibrate with greater amplitude but pitch will not be different as it does not depend on how hard the key is struck.

Question 3.
In SONAR, we us
(a) ultrasonic waves
(b) infrasonic waves
(c) radio waves
(d) audible sound waves
Solution:
(a) Ultrasonic waves are used in SONAR.

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.
Solution:
(c) Sound travels in air if disturbance moves, not the particles of medium.

Question 5.
When we change feeble sound to loud sound we increase its
(a) frequency
(b) amplitude
(c) velocity
(d) wavelength
Solution:
(b) As the loudness and softness of sound or intensity of sound is directly proportional to the square of amplitude of sound wave, therefore, feeble sound can be changed to loud sound by increasing its amplitude.

Question 6.
In the curve half the wavelength is

(a) AB
(b) BD
(c) DE
(d) AE
Solution:
(b) BD represents half the wavelength in the curve.

Question 7.
Earthquake produces which kind of sound before the main shockwave begins
(a) ultrasound
(b) infrasound
(c) audible sound
(d) none of the above
Solution:
(b) Earthquake produces infrasound before the main shock wave begins.

Question 8.
Infrasound can be heard by
(a) dog
(b) bat
(c) rhinoceros
(d) human beings
Solution:
(c) Infrasound can be heard by rhinoceros.

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
Solution:
(c) Frequency of the sitar depends upon tension in string, so, sitarist is adjusting frequency of the sitar string with the frequency of other instruments.

Short Answer Type Questions

Question 10.
Thegiven graph shows the displacement versus time relation for a disturbance travelling with velocity of 1500 m s_1. Calculate the wavelength of the disturbance.

Solution:
From the given graph,
Time period of the disturbance,
T = 2 µ s = 2 x 10^-6s
Velocity of disturbance, v = 1500 m s_l So, wavelength of the disturbance,
λ = vT
= 1500 m s ^-1 x 2 x 10^-6 s = 3 x 10^-3 m

Question 11.
Which of the two graphs (a) and (b) representing the human voice is likely to be the male voice? Give reason for your answer.

Solution:
Usually the male voice has less pitch (or frequency) as compared to female voice.
As the time period of wave represented by graph (a) is more than that of graph (b), so, graph (a) represents lower frequency wave than graph (b) and hence is likely to be the male voice.

Question 12.
A girl is sitting in, the middle of a park of dimension 12 m x 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.
Solution:
If the time gap between the original sound and reflected sound received by the listener is around 0.1 s, only then the echo can be heard.
Speed of sound in air, v=344 m s ^-1

Time gap between the original sound and reflected sound received by the girl
\(=\frac{18}{344} \mathrm{s}-\frac{6}{344} \mathrm{s}=0.035 \mathrm{s}\)
As 0.035 s < 0.1 s, therefore no echo will be heard by the girl.

Question 13.
Why do we hear the sound produced by the humming bees while the sound of vibrations of pendulum is not heard?
Solution:
Sound produced by humming bees by vibrating their wings, lies in the audible range of hearing for average human being which is 20 Hz to 20 kHz, so it can be heard.
While the sound of vibrations of pendulum is less than 20 Hz, so it can not be heard.

Question 14.
If any explosion takes place at the bottom of a lake, what type of shock waves in water will take place?
Solution:
Shock waves in water are longitudinal waves, when any explosion takes place at the bottom of a lake.

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.)
Solution:
Let the distance of thunder cloud from observer be x m.
Now, speed of light = 3 x 108 m s ^-1Speed of sound = 340 m s ^-1
Time taken by the light to reach observer,
\(t_{l}=\frac{x}{3 \times 10^{8}} \mathrm{s}\)
Time taken by the sound to reach observer,
\(t_{s}=\frac{x}{340} \mathrm{s}\)
Given, t₈-t₁=10s

Question 16.
For hearing the loudest ticking sound heard by the ear, find the angle x in the figure.

Solution:
For hearing the loudest ticking sound, angle made by incident sound with normal = angle made by reflected sound wave with normal = x
Now, angle made by incident sound with normal
= 90° – 50° = 40°
∴ x = 40°

Question 17.
Why is the ceiling and wall behind the stage of good conference halls or concert halls made curved?
Solution:
Ceiling and wall behind the stage of good conference halls or concert halls are made curved so that sound after reflection from the wall and ceiling reaches all the parts of the hall.

Long Answer Type Questions

Question 18.
Represent graphically by two separate diagrams in each case
(i) Two sound waves having the same amplitude but different frequencies.
(ii) Two sound waves having the same frequency but different amplitudes.
(iii) Two sound waves having different amplitudes and also different wave lengths.
Solution:
(i) Two sound waves having the same amplitude but different frequencies.

(ii) Two sound waves having the same frequency but different amplitudes.

(iii) Two sound waves having different amplitudes and also different wavelengths.

Question 19.
Establish the relationship between speed of sound, its wavelength and frequency. If velocity of sound in air is 340 ms ^-1, calculate
(i) wavelength when frequency is 256 Hz.
(ii) frequency when wavelength is 0.85 m.
Solution:
Speed of sound is the distance travelled by sound wave per unit time.
Distance travelled by sound wave in periodic time (T) = wavelength (λ) of the sound wave.

thus v = u
Velocity of sound in air, v = 340 m s ^-1
(i) Frequency, u = 256 Hz
∴ Wavelength, λ = \(\frac { v }{ \upsilon } \) = \(\frac { 340 }{ 256 }\) m = 1.33M.
(ii) Wavelength, λ = 0.85 m
∴ Frequency v =\(\frac { v }{ \lambda } \) = \(\frac { 340 }{ 0.85 }\)Hz = 400Hz

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.
Solution:
The points of maximum density (or pressure) and minimum density (or pressure) are also called crests and troughs respectively.

Wavelength (A): The distance between two consecutive compressions or two consecutive rarefactions is called wavelength.
Time period (T): Time period of a wave is defined as the time taken by its two consecutive compressions or rarefactions to cross a fixed point.

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