Maharashtra Board Class 9 Science Solution Chapter 12 – Study of Sound
Balbharati Maharashtra Board Class 9 Science Solution Chapter 12: Study of Sound. Marathi or English Medium Students of Class 9 get here Study of Sound full Exercise Solution.
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Std |
Maharashtra Class 9 |
| Subject |
Science Solution |
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Chapter |
Study of Sound |
1.) Fill in the blanks and explain.
Ans:
a) Sound does not travel through vacuum.
Since, sound waves are longitudinal waves which requires medium for their propagation and hence does not travels through vacuum.
b) The velocity is sound in steel is greater than the velocity of sound in water.
Since, the density of steel is greater than the density of water and hence velocity of sound in steel is greater than the velocity of sound in water.
c) The incidence of thunderstorms in daily life shows that the velocity of sound is less than the velocity of light.
Since, when thunderstorms happens we see first light and then we hear sound from which we conclude that velocity is light is greater than the velocity of sound waves.
d) To discover a sunken ship or objects deep inside the sea SONAR technology is used.
Since, by using this technology we can locate the position or identify the position or location of the animals or objects which are deep inside the sea water.
2.) Explain giving scientific reasons.
a) The roof of a movie theatre and a conference hall is curved.
Ans:
The roof of movie theatre and a conference are made curved because when sound is reflected from curved roof and also from walls, it reach to every part of theatre and conference hall uniformly. So that every person sitting in the theatre or hall can listen sound clearly.
b) The intensity of reverberation is higher in a closed and empty house.
Ans:
In a closed and empty room there will be no objects like furniture and other things. Due to which the sound generated reflects from the walls of the room multiple times and it doesn’t escapes from the room also as room is closed. Because of these continuous reflection of sound without decrease in intensity of sound the reverberations produced are of higher intensity.
c) We cannot hear an echo produced in a classroom.
Ans:
Our classrooms are mostly constructed in a such way that the distance between two walls of the classroom would be less than 17m. When there is sound produced in classroom that reflects from the walls of the classroom and reach to our ear in 0.1 second only. Due to this very small time we can’t distinguish the sound produced originally and the sound reflected. Due to which we can say that in classrooms there will be no echo produced.
3.) Answer the following questions in your own words.
a) What is an echo? What factors are important to get distinct echo?
Ans:
Echo:
Echo is the repeated sound produced due to the continuous reflection of sound from hard surfaces like hill.
Following factors are responsible for producing distinct echo sound:
- The distance between the source from which sound is producing and the hard surface from which sound is reflecting must be minimum 17m.
- And the size of the reflector should be large which is greater than the wavelength of sound produced. Hence, we can hear echo at the top of hills.
b) study the construction of the Golghumat at Vijapur and discuss the reasons for the multiple echos produced there.
Ans:
- The Golghumat at Vijapur has centrally situated dome shaped structure called as ghumat which is standing without any support.
- When the sound is produced inside the dome it will be reflected three to four times within a second. And the sound produced here 3-4 times is nothing but the echo.
- When there will be no people inside the ghumat then there will be 9-10 times echo can be produced.
- Hence, structure so called ghumat is responsible for producing echo many times in Golghumat at Vijapur.
c.) What should be the dimensions and the shape of classrooms so that no echo can be produced there?
Ans:
The dimensions of classroom should be made such that we can’t hear an echo is the classroom should be square shaped and the distance between walls to be maintained less than 17.2 m.
And to hear the sound produced uniformly and clearly to every person sitting in the classroom the ceiling should be of curved shaped also.
4.) Where and why sound absorbing materials are used?
Ans:
The sound absorbing materials are used in constructing the roofs and walls of conference halls, auditorium halls and theatre so that the sound reflected from the roof and walls will be absorbed by the sound absorbing materials and hence the intensity of reverberation produced will be decreased.
5.) Solve the following examples:
a) The speed of sound in air at 0°C is 332m/s. If it increases at the rate of 0.6m/s per degree, what will be the temperature when the velocity has increased to 344m/s.
Ans:
Let T be the temperature at which velocity of sound is 344m/s.
Then the velocity of sound increased from 0°C to T°C would be = 344 – 332= 12m/s
But, given that the velocity of sound increases by 0.6m/s per degree rise in temperature.
Hence, the temperature required to meet the extra velocity 12m/s would be (12/0.6)= 20°C
Thus, the velocity of sound becomes 344m/s when the temperature is 20°C.
b) Nita heard the sound of lightning after 4seconds of seeing it. What was the distance of lightning from her? The velocity of sound in air is 340m/s.
Ans:
Given that velocity is sound in air is 340m/s.
And Nita heard sound of lightning after 4seconds.
Hence, the distance of lightning from Nita become= 340*4= 1360m
Thus, the distance of lightning from Nita would be 1360m.
c) Sunil is standing between two parallel walls. The wall closest to him is at a distance of 660m. If he shouts, he hears the first echo after 4s and another after another 2s.
1) What is the velocity of sound in air?
2) What is the distance between the two walls?
Ans:
Given that,
The distance of wall closest to him is 660 m.
And first echo will be heard after 4 seconds.
1)
Thus, distance travelled by the sound in 4second would be = 660*2
Hence, velocity of sound in air = 660*2/4= 330m/s
Hence, the velocity of sound in air would be 330m/s.
2)
The time required to hear second echo= 2+4= 6 seconds
Let distance travelled in 6s would be= 2*X
And velocity of sound in air is 330m/s.
Hence, we can write,
2*X= 330*6
Thus, X= 990m
Hence, the distance between the two walls is= 990+660= 1650m
d) Hydrogen gas is filled in two identical bottles, A and B at the same temperature. The mass of hydrogen in the two bottles is 12g and 48g respectively. In which bottle sound will travel faster? How many times as fast as the other?
Ans:
Given, mass of bottle A= m = 12 g
mass of bottle B = M = 40 g
Let us suppose speed of sound in bottle A = v
And speed of sound in bottle B = V
As we know that speed of sound= 1/√d
Where d- density at constant temperature
But, density = mass/ volume
d =m/v
Where m – mass
V – volume
Speed of sound = 1/√d
Speed of sound = 1/√(m/v)
Speed of sound = √v/√m
Firstly we calculate speed of sound in bottle A= √v/√m…
But volume is constant
Sound in bottle A = 1/√m
Sound in bottle A = 1/√12
Similarly we calculate Speed of sound in bottle B= 1/√48
(Speed of sound in bottle A)/(Speed of sound in bottle B) = √48/√12
v/V= √4
= 2
v = 2V
Thus Speed of sound in bottle A is 2 times greater than speed of sound in bottle B.
e) Helium gas is filled in two identical bottles A and B. The mass of the gas in two bottles is 10g and 40g respectively. If the speed of sound is the same in both the bottles, what conclusion will you draw?
Ans:
Given that,
Mass of gas in bottle = m = 10 g
Mass of gas in bottle = M= 40 g
speed of sound in bottle A =speed of sound in bottle B = V
Temperature of bottle A = t
Temperature of bottle B = T
Density of bottle A = density of bottle B = D
As we know that,
(Speed of sound in bottle A)/(Speed of sound in bottle B)= [(√M/D)/ (√m/D)] × (√t/√T)
1= (√M/√m)× (√t/√T)
(√T/√t) = (√M/√m)
(√T/√t) = (√40/√10)
(√T/√t) = √4
Taking square on both sides,
T/t = 4
T = 4t
Thus Temperature of bottle B is 4 time greater than Temperature of bottle A.
