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## Sunday, November 22, 2015

### 9th Sound Test Paper Solved

1.A submarine enters a sonar pulse, which returns from an under water cliff in 1.02 s. If the speed of sound in salt water is 1531 m/s, how far away is the cliff?
Ans: t = 1.02s, v = 1531m/s Now d=s x t Distance travelled by sonar pulse = 2d = 2x1531x1.02 = 1561.62m

Distance of the cliff = d/2= 1561.62/2 = 780.81m

2. If the tension in the wire is increased four times how will the velocity of wave in a string varies?
Answer: velocity of the wave in string is directly proportional to the square root of the tension thus if tension is increased 4 times the velocity will be doubled.

3. What is Echo. Explain the conditions that have to be satisfied to hear an echo?
Ans: reflection of sound wave from a large obstacle is called an echo.

The most important condition for hearing an echo is that the reflected sound should reach the ear only after a lapse of at least 0.1 second after the original sound dies off and the obstacle is at least at a distance of 17 m

4. Why do echoes produced in an empty auditorium usually decrease when it is full of audience?
Ans: When the hall is empty there is no obstacles in between to reflect the sound other than the walls. When the hall is full of audiences the sound produced undergoes multiple reflection from the people and so it overlaps with the sound produced. Hence the listener is not able to distinguish between the original sound and the echo.

5. A girl claps and hears the echo after reflection from cliff which is 660 m away. If the velocity of sound is 330 m s-1, calculate the time taken for hearing the echo.

Ans: v x t = 2 d Þ t = 2d/v = 2x660/300 = 4s

6. Explain how is the principle of echo used by the dolphin to locate small fish as its prey.
Ans: Dolphins are aquatic animals which send out ultrasonic sound to communicate with each other. They have a sound sensing system which enables them to find animals under water with great accuracy due to the echo of the ultrasonic sound produced by them.

7. Explain by some experiment that sound waves require medium for their propagation.
Ans: An electric bell is suspended inside an airtight glass bell jar connected to a vacuum pump. As the electric bell circuit is completed, the sound is heard. Now if the air is slowly removed from the bell jar by using a vacuum pump, the intensity of sound goes on decreasing and finally no sound is heard when all the air is drawn out. We would be seeing the hammer striking the gong repeatedly. This clearly proves that sound requires a material for its propagation.

8. Distinguish between loudness and intensity of sound.
Ans: Intensity depends on the energy per unit area of the wave and it is independent of the response of the ear, but the loudness depends on energy as well as on the response of the ear.

9. Give two practical applications of the reflection of sound waves.

Ans: (i) In stethoscope the sound of patient’s heartbeat reaches the doctor’s ears by multiple reflections in the tubes.

(ii) Megaphones are designed to send sound waves in particular direction are based on the reflection of sound.

10. Why are longitudinal waves called pressure waves?

Ans: Sound waves travels in the form of compression and rarefactions, which involve change in pressure, and volume of the air. Thus they are called pressure waves.

11. Sound travels faster on a rainy day than on a dry day. Why?
Ans: Sound travels faster on rainy day because the velocity of sound increases with increase in humidity. On rainy day humidity is more thus velocity of sound is also more.

12. How moths of certain families are able to escape capture?
Ans: Moths of certain families can hear high frequency sounds (squeaks) of bat as they have sensitive hearing equipment. Thus they get to know when a bat is near by and hence able to escape its capture

Extra score