From a height of 2 m a drop of water of radius `2xx10^(-3)` m fall and produces a sound. The sound produced can be heard upto a distance of 20 m. If the gravitational energy is converted into sound energy in 0.5 s, then calculate the intensity at a distance of 20 m.
A
`2xx10^(-7)W//m^(2)`
B
`2.6xx10^(-6)W//m^(2)`
C
`2.6xx10^(-7)W//m^(2)`
D
`3xx10^(-7)W//m^(2)`
Text Solution
Verified by Experts
The correct Answer is:
C
Change in garviational energy = mgh `=(4/3pir^(3))` pgh jouic Hence, power `=(4/3pir^(3)pgh)/(t)\` watt and intensity `=(4/3pir^(3)pgh)/((4piR^(2))t)=2.6xx10^(-7)W//m^(2)`
Topper's Solved these Questions
SOUND WAVES
BITSAT GUIDE|Exercise BITSAT Archives|1 Videos
SOLVED PAPER 2018
BITSAT GUIDE|Exercise Part-I (Physics)|40 Videos
UNIVERSE
BITSAT GUIDE|Exercise All Questions|30 Videos
Similar Questions
Explore conceptually related problems
At a distance of 5.0 m from a point sound source, the sound intensity level is 110 dB. At what distance is the intensity level 95 dB?
A drop of water 2 mm in diameter falling from a height of 50 cm in a bucket generates sound which can be heard from 5 m distance . Take all gravitational energy difference equal to sound energy , the transformation being spread in time over 0.2 sec, deduce the average intensity. Take g=10 ms^(-2)
Loudness of sound from an isotropic point source at a distance of 10m is 20dB . If the distance at which it is not heard is 10^(K) in meters find K .
The intensity of sound from a radio at a distance of 2 metres from its speaker is 1xx10^(-2) mu W//m^(2) . The intensity at a distance of 10 meters would be
The sound produced by a siren given by a company reaches a person on the road at a distance of 110 m in _______ s. (Velocity of sound in air is 330 ms^(-1) )
The echo of a sound is heard after 5 seconds. If the speed of sound in air be 342m/s, calculate the distance of the reflecting surface.
