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

#### Page No 137:

#### Question 1:

a. Sound does not travel through ............

#### Answer:

__vacuum__.

__greater__than the velocity of sound in water.

__thunderstorm__in daily life shows that the velocity of sound is less than the velocity of light.

__SONAR__technology is used.

#### Page No 137:

#### Question 2:

#### Answer:

a. The roof of a movie theatre and a conference hall is curved so that the sound after reflection from the ceiling reaches all the parts of the hall uniformly.

b. In an empty house, there is no sound absorbers present such as furnitures, curtains etc. Also, if the room is closed, the sound cannot escape the room. Thus, the sound goes through multiple reflections from the rigid walls of a closed and empty room without the reflections of the sound being absorbed. Hence, the intensity of reflections do not fade which would have faded if the room was not empty (has sound absorbers). Due to this the intensity of reverberation is higher in a closed and empty room.

c. The classrooms are designed in such a way that the distance between the walls are less than 17.2 m. Due to this, the reflections of sound from to walls or echos reach our ears before 0.1 s. Because of this, we are not able to distinguish between the original sound and the echo produced in the classroom.

#### Page No 137:

#### Question 3:

a. What is an echo? What factors are important to get a distinct echo?

#### Answer:

a. The repetition of sound caused by its reflection off a hard surface is known as echo. Factors important to get a distinct echo are:

- The minimum distance between the source and reflector of sound should be 17 m.
- The size of reflector must be large enough as compared to the wavelength of the sound wave.

c. For no echo to be produced in a classroom, the classroom should be square shaped with distance between the walls to be less than 17.2 m. Also, the ceiling should be curved shaped so that the reflection of sound wave is uniform throughout the room.

#### Page No 137:

#### Question 4:

Where and why are sound absorbing materials used?

#### Answer:

The sound absorbing materials are used on the roofs and walls of auditoriums, concert halls, theatres, etc. as well as upholsteries choosen for these rooms are made of sound absorbing materials. In these places, excessive reverberation is highly undesirable. Due to the presence of sound absorbing materials, the sound reflected from the rigid surfaces is absorbed by these materials and hence reverberation is subdued.

#### Page No 137:

#### Question 5:

a. The speed of sound in air at 0

^{o}C is 332 m/s. If it increases at the rate of 0.6 m/s per degree, what will be the temperature when the velocity has increased to 344 m/s?

#### Answer:

a. Let the temperature at which the velocity of air becomes 344 m/s be *T*^{o}C.

The increase in velocity of sound in air from 0^{o}C to *T*^{o}C = 344$-$332 = 12 m/s

Given,

The velocity of sound in air increases by 0.6 m/s for 1^{o}C rise in temperature.

Therefore, for 1 m/s increase in velocity of sound in air, the temperature should rise by $\frac{1}{0.6}{}^{\mathrm{o}}\mathrm{C}$.

For 12 m/s increase in velocity of sound in air, the temperature should rise by = $\frac{12}{0.6}{}^{\mathrm{o}}\mathrm{C}=20{}^{\mathrm{o}}\mathrm{C}$

Thus, the temperature at which the velocity of air becomes 344 m/s = *T*^{o}C = 20^{o}C$-$ 0^{o}C = 20^{o}C

b. Let the distance between Nita and the source of lightning be *x*.

Speed of sound in air = 340 m/s

Time taken by sound to reach Nita = 4 s

*x* = 340$\times 4$ = 1360 m

c.

1. Time to hear first echo = 4 s

Distance travelled by sound in these 4 s = 360$\times 2$ m

Speed of sound = $\frac{360\times 2}{4}=180\mathrm{m}/\mathrm{s}$

2. Time taken to hear second echo = 4 s + 2 s = 6 s

Distance travelled in these 6 s = 2*x*

Speed of sound in air = 180 m/s

Therefore, 2*x = *180* *$\times 6$

*x* = 540 m

Distance between the walls = 540 + 360 = 900 m

d. The velocity of sound in gas is related to density of gas as

$v\propto \frac{1}{\sqrt{\rho}}$ .....(i)

and the velocity of sound in gas is related to temperature of gas as

$v\propto \sqrt{T}$ .....(ii)

Combining (i) and (ii), we get

$v\propto \frac{\sqrt{T}}{\sqrt{\rho}}$

Now, the two bottles given are identical i.e. the volumes of gases are same. Let the volume of the bottle be *V*. Let *M*_{1} and *M*_{2} be the masses of the gases in bottles A and B, respectively and *v*_{1} and *v*_{2} be the velocity of the sound in the two bottles, respectively. Since, the temperature of gas in both the bottles are same, let that common temperature be* T*. Therefore,

$\frac{{v}_{1}}{{v}_{2}}=\frac{\sqrt{{\displaystyle \frac{{M}_{2}}{V}}}\sqrt{T}}{\sqrt{{\displaystyle \frac{{M}_{1}}{V}}}\sqrt{T}}=\frac{{\displaystyle \sqrt{48}}}{{\displaystyle \sqrt{12}}}=\sqrt{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{1}=2{v}_{2}$

Thus, the sound travels 2 times faster in bottle B as compared to bottle A.

e. Mass of the gas in bottle A = 10 g

Mass of gas in bottle B = 40 g

The velocity of sound in gas is related to density of gas as

$v\propto \frac{1}{\sqrt{\rho}}$ .....(i)

and the velocity of sound in gas is related to temperature of gas as

$v\propto \sqrt{T}$ .....(ii)

Combining (i) and (ii), we get

$v\propto \frac{\sqrt{T}}{\sqrt{\rho}}$

The bottle are identical, this means the volumes of the gases are equal. Let the volume of the bottle be *V*. Let *M*_{1} and *M*_{2} be the masses of gases in bottles A and B, respectively and *v*_{1} and *v*_{2} be the velocity of the sound in the two bottles, respectively. Also, let *T*_{1} and *T*_{2} be their respective temperatures. Therefore,

$\frac{{v}_{1}}{{v}_{2}}=\frac{\sqrt{{\displaystyle \frac{{M}_{2}}{V}}}\sqrt{{T}_{1}}}{\sqrt{{\displaystyle \frac{{M}_{1}}{V}}}\sqrt{{T}_{2}}}$

Now given that, *v*_{1} = *v*_{2}

$\Rightarrow \sqrt{{M}_{1}}\sqrt{{T}_{2}}=\sqrt{{M}_{2}}\sqrt{{T}_{1}}\phantom{\rule{0ex}{0ex}}\mathrm{or},{T}_{2}=\frac{{M}_{2}{T}_{1}}{{M}_{2}}\phantom{\rule{0ex}{0ex}}\mathrm{Given},{M}_{1}=10\mathrm{g},{M}_{2}=40\mathrm{g}\phantom{\rule{0ex}{0ex}}\Rightarrow {T}_{2}=\frac{40{T}_{1}}{10}=4{T}_{1}\phantom{\rule{0ex}{0ex}}$

Thus, it can be concluded that the temperature of bottle B is 4 times the temperature of A.

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