Question
If a person drops a pebble in a well of 100 m depth, after how much time (in seconds) will he be able to hear the sound?
Given:
Acceleration due to gravity, g=10 ms−2
Speed of sound in air, v=343 ms−1
Given:
Acceleration due to gravity, g=10 ms−2
Speed of sound in air, v=343 ms−1
Answer: Option A
:
A
The question can be divided into two parts, where in the first part, thetime taken by the pebble to reach the bottom is calculated and in thesecond part, the time taken by the sound wave to reach the top is calculated.
Given:
Initial velocity,u=0ms−1
Depth of the well, s=100m
Acceleration, a=10ms−2
Speed of sound, v=343ms−1
Let t1 be the time taken by the pebble to hit the bottom.
From second equation of motion,
s=ut1+12at21
100=(0×t1)+(12×(10)×t21)
100=5t21
⇒t1=√20s
We know, time=distancespeed
Let t2 be the time taken for the sound wave to reach from the bottom to the top of the well.
t2=sv=100343≈0.3s
Total time,
t=t1+t2
=√20+0.3s.
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:
A
The question can be divided into two parts, where in the first part, thetime taken by the pebble to reach the bottom is calculated and in thesecond part, the time taken by the sound wave to reach the top is calculated.
Given:
Initial velocity,u=0ms−1
Depth of the well, s=100m
Acceleration, a=10ms−2
Speed of sound, v=343ms−1
Let t1 be the time taken by the pebble to hit the bottom.
From second equation of motion,
s=ut1+12at21
100=(0×t1)+(12×(10)×t21)
100=5t21
⇒t1=√20s
We know, time=distancespeed
Let t2 be the time taken for the sound wave to reach from the bottom to the top of the well.
t2=sv=100343≈0.3s
Total time,
t=t1+t2
=√20+0.3s.
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