Science Ncert Exemplar 2019 Solutions for Class 9 Science Chapter 12 Sound are provided here with simple step-by-step explanations. These solutions for Sound are extremely popular among class 9 students for Science Sound Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Science Ncert Exemplar 2019 Book of class 9 Science Chapter 12 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Science Ncert Exemplar 2019 Solutions. All Science Ncert Exemplar 2019 Solutions for class 9 Science are prepared by experts and are 100% accurate.

Page No 71:

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 sound produced by single frequency is called tone and if a sound is produced due to multiple frequencies then it is called Note.

So, note is a sound of mixture of several frequencies. 

Hence, the correct answer is option (a).

Page No 71:

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:

When the key of a mechanical piano is struck harder, sound will be louder as it will vibrate with greater amplitude but the frequency of the sound remains same. Pitch of the sound depends on frequency.
So, it can be said the pitch does not depend on how hard the key is struck. 

Thus, It can be concluded that sound will be louder but the pitch will not be different.


Hence, the correct answer is option (a).

Page No 71:

Question 3:

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

Answer:

Sound Navigation and Ranging (SONAR) is a technology used to detect obstacles in a path. This technique is used by bats and whales.
SONAR uses ultrasonic waves for its operation. This is because ultrasonic waves have slightly larger frequency than audible range. 
Thus, they have higher energy to penetrate obstacles. 
Using ultrasonic waves in submarines etc. also ensures not mixing signals with other noises like noise of engines etc.

 

Hence, the correct answer will be option (a)

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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:

In case of the sound waves particle vibrates about their mean position and the disturbance travels through the medium.

Hence, the correct answer is option (c).

 

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Question 5:

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

Answer:

Intensity of the sound depends only on the amplitude of the vibration 

IntensityAmplitude2

Feeble sound can be changed to loud sound only by increasing the amplitude of the sound waves.

Hence, the correct answer is option (b).

Page No 71:

Question 6:

In the curve Figure half the wavelength is

(a) A B
(b) B D
(c) D E
(d) A E

Answer:

BD represents the half wavelength

Hence, the correct answer is option (b).



Page No 72:

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:

Earthquake is a movement of tremor of the earth's crust.  It produces infrasound before the mainshock wave begins. Infrasound is low-frequency sound less than 20 Hz. It can travel hundreds of km through the earth's surface. So by detecting this sound, we can predict an earthquake. ​
Infrasounds are not audible to human ear. But many animals and birds are able to hear it and they become alert.

Hence, the correct answer is option (b).

Page No 72:

Question 8:

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

Answer:

Infrasound is the sound with less frequency than audible range of humans.
Rhinos have extremely good ears, picking up infra-sound far deeper than the range of human hearing. Rhinos, can hear down to a frequency of four hertz,
Hence, the correct answer is option (c).

Page No 72:

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:

Frequency of the sitar depends upon tension in string, So a sitarist tries to adjust the tension and tries to match the frequency of the sitar with other instruments.

Hence, the correct answer is option (c).

 

Page No 72:

Question 10:

The given graph Figure shows the displacement versus time relation for a disturbance travelling with velocity of 1500 m s-1. Calculate the wavelength of the disturbance.

Answer:

Wavelength of the disturbance or wave can be found by using the formula λ = velocity(v) frequency (f) 

From the figure the time period T of the wave is 2 µ s = 2×10-6 s

Velocity of the disturbance is 1500 m s-1 .

 λ = velocity(v) frequency (f) = velocity(v) ×Time period (T) = 1500×2×10-6 = 3×10-3 m​

Page No 72:

Question 11:

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

Answer:

Male voice has less pitch (or frequency) as compared to female voice.

As we can see from the graph, ​​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.



Page No 73:

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:

Sound of the cracker will travel towards the building and get reflected from there, So the total distance travelled by the sound so that the eco is there will be = 12 + 6 = 18 m.

Time between the original and the reflected sound is given by, t = 18 344-6344= 0.035 s
 

Due to the persistence of sound one can only hear two sound distinctly if the time interval between them is greater than 0.10s.

But 0.035 is less than 0.10 s.
So the girl will not be able to hear the echo of the sound.

 

 

Page No 73:

Question 13:

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

Answer:

Sound produced by humming bees due to the vibrations  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.

Page No 73:

Question 14:

If any explosion takes place at the bottom of a lake, what type of shock waves in water will take place?

Answer:

Longitudinal waves, are produced when any explosion takes place at the bottom of a lake. These generated longitudinal waves are a type of shock waves.

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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:

Time taken by the sound wave to reach us = 10 s

Speed of sound = 340 ms-1 
 
Distance travelled by the sound wave = 340×10 = 3400 m = 3.4 km.

Hence, the approximate distance of the thunder cloud is 3.4 km.

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Question 16:

For hearing the loudest ticking sound heard by the ear, find the angle x in the Figure.

Answer:

From the figure the angle made by the incident wave with the normal is given by 90- 50= 40 

For hearing the largest ticking sound the angle of incidence should be equal to the angle of reflection of the sound wave i.e. the angle of reflection x  should be equal to 40.

 

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Question 17:

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

Answer:

The ceiling of concert halls, conference halls, and cinema halls are curved so that sound after reflection reaches all corners of the hall uniformly.

Page No 73:

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:


Page No 73:

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:

Speed of sound = frequency of the sound × Wavelength of the sound.

(i)  frequency f = 256 Hz

  Wavelength λ = speed of Sound frequency of sound= 340256=1.33 m

(ii) Wavelength = 0.85 m

   frequency = Speed Wavelength=3400.85= 400 s-1 or 400 Hz
 

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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:

The wavelength is given by the distance between two successive crests or troughs.
The time period is the time required for one complete oscillation.



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