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Question

A boy stands between the two walls as shown below and he claps his hands.


If ‘x’ is less than 60m and the time between the first and the second echo is 0.25s, then calculate the distance ‘x’.
Velocity of sound in air is 344 ms1.

A
36m
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B
34m
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C
40m
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D
17m
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Solution

The correct option is D 17m
Given:
Speed of sound, v= 344 ms1
Time difference between the first and the second echo, Δt=0.25s.

Total distance travelled by sound in 0.25 s is
Δs=vΔt
Δs=344×0.25=86m

The distance of boy from the right wall is 60 m, the total distance travelled by the sound in the second echo is:
s2=120 m

The distance of the boy from the left wall is x.
The total distance travelled by the first echo is: s1=2x m.

Total distance travelled by sound in 0.25 s is equal to the difference in the distance covered by the second echo and the first echo.
Δs=s2s1
86 m=120 m2x
x=17 m

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