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 0.3 s √20 s  20 s √20+0.3 s

Solution

## The correct option is D √20+0.3 s The question can be divided into two parts, where in the first part, the time taken by the pebble to reach the bottom is calculated and in the second part, the time taken by the sound wave to reach the top is calculated. Given:  Initial velocity, u=0 ms−1 Depth of the well, s=100 m Acceleration, a=10 ms−2 Speed of sound, v=343 ms−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=√20 s 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≈0.3 s Total time, t=t1+t2   =√20+0.3 s.

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