# NCERT Exemplar Solutions for Class 9 Science Chapter 12 Sound (MCQ, SAQ and LAQ)

 Chapter Name NCERT Exemplar Solutions for Class 9 Science Ch 12 Sound Topics Covered Objective Type Questions (MCQ's)Short Answer Type QuestionsLong Answer Type Questions Related Study NCERT Solutions for Class 9 ScienceNCERT Revision Notes for Class 9 ScienceImportant Questions for Class 9 ScienceMCQ for Class 9 ScienceNCERT Exemplar Questions For Class 9 Science

### Objective Type Questions for Sound

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.

3. In SONAR, we use :
(a) ultrasound waves
(b) infrasonic 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.

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

Solution

(b) amplitude

6. In the half curve the wavelength is :

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

Solution

(b) 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

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

Solution

(c) Rhinoceros

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.

### Short Answer Questions for Sound

10. The given graph (fig.) 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 graph
Time period, T = 2 × 10-6 s
Frequency, Î½ = 1/T = 5 × 105 Hz
Wavelength, Î» = Velocity(v)/Frequency(Î½)
= 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 male voice has less pitch (or frequency) as compared to female. Thus, graph (a) represents the 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 time gap between the reflected sound and original sound received by the listener is around 0.1 s, only then the echo can be heard. The minimum distance travelled by the reflected sound wave for distinctly listening the echo = velocity of sound × time interval
Then, 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 sound by vibrating their wings which is in the audible range. In case of pendulum the frequency is below 20 Hz which does not come in the audible range.

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.

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
Distance = 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

From the diagram, ∠i = ∠r
So, x = 90° − ∠r = 90° − 50° = 40°
For hearing the loudest ticking sound heard by the ear the angle is 40° .

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

Solution

Ceiling and walls are made curved so that sound after reflection reaches the target audience.

### Long Answer Questions for Sound

18. 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 time taken to travel the distance between any two consecutive compression or refraction from a fixed point.

19. 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

(a) Same amplitude but different frequency

(b) Same frequency but different amplitude
(c) different amplitude and different wavelengths

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

(i) Derivation of formula, v = Î½ Î»
340 = 256 Î»
Î» = 340/256 = 1.33 m

(ii) Frequency = speed/wavelength
= 340/0.85
= 400 Hz