## Thursday, 4 February 2016

### TEXTBOOK ANSWERS AND SOLUTIONS OF CBSE CLASS IX SCIENCE Chapter 12 Sound

1) What is sound and how is it produced?

ANSWER:-Sound is a form of energy which gives the sensation of hearing. It is produced by the vibrations caused in air by vibrating objects.

2) Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.

When a vibrating body moves forward, it creates a region of high pressure in its vicinity. This region of high pressure is known as compressions. When it moves backward, it creates a region of low pressure in its vicinity. This region is known as a rarefaction. As the body continues to move forward and backwards, it produces a series of compressions and rarefactions.

3) Cite an experiment to show that sound needs a material medium for its propagation.

ANSWER:-Take an electric bell and an air tight glass bell jar connected to a vacuum pump. Suspend the bell inside the jar, and press the switch of the bell. You will be able to hear the bell ring. Now pump out the air from the glass jar. The sound of the bell will become fainter and after some time, the sound will not be heard. This is so because almost all air has been pumped out. This shows that sound needs a material medium to travel.

4) Why is sound wave called a longitudinal wave?

ANSWER:-Sound wave is called longitudinal wave because it is produced by compression  and rarefaction in the air.

5) Which characteristics of the sound helps you to identify your friend by his voice while sitting with others in a dark room?

ANSWER:-The quality or timber of sound enables us to identify our friend by his voice.

6) Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?

ANSWER:-Sound of thunder takes more time to reach the Earth as compared to light. Hence, a flash is seen before we hear a thunder.

7). A person has a hearing range from 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies? Take the speed of sound in air as 344 m s 1 .

Frequencyv = λ x ν
Speed of sound in air = 344 m/
s (Given)
(i) For, ν= 20 Hz
λ1= v/ν = 344/20 = 17.2 m
(ii) For, ν= 20000 Hz
λ2= v/ν = 344/20000 = 0.172 m

8) Two children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminium to reach the second child.

ANSWER:-Velocity of sound in air= 346m/s
Velocity of sound wave in
Aluminium= 6420 m/s
Let length of rode be 1
Time taken for sound wave in
Air, t 1= 1 / Velocity in air
Time taken for sound wave in
Aluminium, t2= 1 / Velocity in aluminium
Therefore, t1 / t2 = Velocity in aluminium / Velocity in
air =6420 / 346 = 18.55 : 1

9. The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?

This means the source of sound vibrates 100 times in one second.
Therefore, number of vibrations in 1 minute, i.e. in 60 seconds
= 100 x 60 = 6000 times.

10) Does sound follow the same laws of reflection as light does? Explain.

ANSWER:-incident sound wave and the reflected sound wave make the same angle with the normal to the surface at the point of incidence. Also, the incident sound wave, the reflected sound wave, and the normal to the point of incidence all lie in the same plane. Therefore they follow the same laws.

11) When a sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source of sound production remains the same. Do you hear echo sound on a hotter day?

ANSWER:-Time Taken= Total Distance / Velocity
On a hotter day, the velocity of sound is more. If the time taken by echo is less than 0.1 sec it will not be heard.

12) Give two practical applications of reflection of sound waves.

ANSWER:-Two practical applications of reflection of sound waves are:
a) Working of a stethoscope is also based on reflection of sound.
b) Used to measure the distance and speed of underwater objects.

13) A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top? Given, g = 10 m s 2 and speed of sound = 340 m s 1.

ANSWER:-Height of the tower, s = 500 m
Velocity of sound, v = 340 m s1
Acceleration due to gravity, g = 10 m s 2
Initial velocity of the stone, u = 0 (since the stone is initially at rest)
Time taken by the stone to fall to the base of the tower, t1

Time taken by the sound to reach the top from the base of the tower, t 2= 500 / 340 =1.47 s
Therefore, the splash is heard at the top after time, t
Where, t= t 1 + t 2 = 10 + 1.47 = 11.47 s.

14. A sound wave travels at a speed of 339 m s 1. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?

ANSWER:-Speed of sound, v= 339 m s - 1
Wavelength of sound, λ= 1.5cm = 0.015 m
Speed of sound = Wavelength
x Frequency= λ x v
v= v / λ = 339 / 0.015 = 22600 Hz
Since the frequency of the given sound is more than 20,000 Hz, it is not audible.

15) What is reverberation? How can it be reduced?

ANSWER:-The repeated multiple reflections of sound in any big enclosed space is known as reverberation. The reverberation can be reduced by covering the ceiling and walls of the enclosed space with sound absorbing materials

16) What is loudness of sound? What factors does it depend on?

ANSWER:-The effect produced in the brain by the sound of different frequencies is called loudness of sound. Loudness depends on the amplitude of vibrations

17) Explain how bats use ultrasound to catch a prey.

ANSWER:-Bats produce high-pitched ultrasonic squeaks. These high-pitched squeaks are reflected by objects such as preys and returned to the bat's ear

18) How is ultrasound used for cleaning?

ANSWER:-Objects to be cleansed are put in a cleaning solution and ultrasonic sound waves are passed through that solution. The high frequency of these ultrasound waves detaches the dirt from the objects.

19) A sonar device on a submarine sends out a signal and receives an echo 5 s later.
Calculate the speed of sound in water if the distance of the object from the submarine is
3625 m.

Distance of the object from the submarine, d = 3625 m
Total distance travelled by the sonar waves during the transmission and reception in water= 2d
Velocity of sound in water, v= 2d / t = 2 x 3625 / 5 = 1450ms-1.

21. Explain how defects in a metal block can be detected using ultrasound.