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NCERT Exemplar Chapter 12 Sound Class 9 Science Solutions

NCERT Exemplar Class 9 Science Chapter 12 Sound Solutions

NCERT Exemplar Solutions for Class 9 Science Chapter 12 Sound covers all the important questions and answers as well as advanced level questions. It helps in learning about the production of sound, vibrations, production of human voice, propagation of sound, medium, characteristics of a sound waves, longitudinal waves, Reflection of sound, echo, reverberation, multiple reflections of sound.

The NCERT Exemplar solutions for class 9 science is very important in the examination. NCERT Exemplar Solutions for Class 9 Science Chapter 12 Sound is provided by our experts. They prepared the best solutions which help the students in understanding the solutions in an easy way. This chapters also covers the other topics like applications of ultrasound, ultrasound for medical use, structure of the human ear, audible frequencies, concept and diagram of the human ear.


Chapter Name

Chapter 12 Sound

Book Title

NCERT Exemplar for Class 9 Science

Related Study

  • NCERT Solutions for Class 9 Science Chapter 12 Sound
  • Revision Notes for Class 9 Science Chapter 12 Sound
  • MCQ for Class 9 Science Chapter 12 Sound
  • Important Questions for Class 9 Science Chapter 12 Sound

Topics Covered

  • MCQ
  • Short Answers Questions
  • Long Answers Questions

NCERT Exemplar Solutions for Chapter 12 Sound Class 9 Science

Multiple Choice Questions

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

(c) of a single frequency.


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) sound will be louder but pitch will not be different.
Pitch depends on the frequency and loudness depends on force by which the key is pressed.


3. In SONAR, we use :
(a) ultrasound waves
(b) infrasonic waves
(c) radio waves
(d) audible sound waves

Solution

(a) ultrasound waves


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) disturbance moves.
Sound waves propagate by vibrating. Disturbances created by vibration of particles move from one place to another.


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

Solution

(b) amplitude
Loudness of sound is proportional to amplitude. When the amplitude of a feeble sound increases, it changes to loud sound.


6. In the curve, half the wavelength is:

(a) AB
(b) BD
(c) DE
(d) AE

Solution

(b) BD
Wavelength is the distance between two consecutive troughs. In the graph half the wavelength is BD.


7. Earthquake produces which kind of sound before the main shock wave begins?
(a) Ultrasound
(b) Infrasonic
(c) Audible
(d) Inaudible

Solution

(b) Infrasonic
Few animals sense the earthquake and start behaving abnormally before earthquake due to infrared rays.


8. Infrasound can be heard by :
(a) Dog
(b) Bat
(c) Rhinoceros
(d) Human being

Solution

(c) Rhinoceros
Infrasound has a frequency less than 20 Hz. Rhinoceroses communicate using infrasound waves of frequency of 5 Hz.


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 string with the frequency of other musical instruments.

Before beginning to play instruments, the artists adjust the frequencies. This is because the musical instruments should be tuned in with other musical instruments to produce pleasant music.


Short Answer Questions

10. The given graph (fig.) shows the displacement versus time relation for a disturbance travelling with velocity of 1500 ms-1. Calculate the wavelength of the disturbance.

Solution

From the graph
Time period, T = 2 × 10-6 s
Frequency, ν = 1/T = 5×105 Hz
v = 1500 ms-1

Wavelength, λ = Velocity(v)/Frequency(ν)
= 1500/(5× 105
)
= 3 × 10-3 m


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

Solution

Usually, the pitch of a male voice is lighter than the pitch of female. Thus, the graph a represents male voice.


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.

Solution

If the gap between the original sound and reflected sound received by the listener is around 0.1 second, only then the echo can be heard. The minimum distance travelled by the reflected sound wave for distinctly listening the echo is
Distance = Sound Velocity × time interval
= 344×0.1
= 34.4 m
In this case, the distance travelled by the sound reflected from the building and then reaching to the girl will be (6 + 6) = 12 m, which is much smaller than the required distance. Therefore, no echo can be heard.


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

Solution

Humming bees produce the sound by beating their wings. The frequency of this sound lies in the range of 20Hz to 20000 Hz which is audible. On the other hand, pendulum produces sound less than 20 Hz. This is below the audible range. This is the reason why we don’t hear the sound of pendulum vibrations.


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

Solution

Longitudinal waves will take place in water if any explosion takes place at the bottom of a lake.


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

Solution

Speed of sound = 340 ms-1
time = 10 s
We know,
speed = distance/time
Therefore,
distance = speed × time
= 340 × 10
= 3400 m


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

Solution

Angle of incidence is always equal to the angle of reflection.
Angle of incidence = 90° -50° = 40°
Angle of reflection= angle of incidence =40°
Hence angle x is 40°.


17. Why is 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 curved. This ensures that the reflected sound reaches to the audience equally.


Long Answer Questions

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.

Solution


19. 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

Wavelength is the distance between two consecutive compression and rarefaction. Time period is the time taken to travel the distance between any two consecutive compression or refraction from a fixed point.


20. 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

Relationship between sound speed, wavelength and frequency:

Speed, v= wavelength × frequency

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