Last updated on June 17th, 2019 at 02:47 pm

## Problem Statement

**47) A pebble is dropped freely in a
well from its top. It takes 20 s for the pebble to reach the water surface in
the well. Taking g = 10 m/s^2 and speed of sound = 330 m/s, find: (i)
the depth of water surface, and (ii) the time when echo is heard after the
pebble is dropped. **

## Solution

** **Say, the depth of water surface in the well

= the distance between the point where from the pebble is released and the surface of water in the well = H

It takes 20 s for the pebble to reach the water surface in the well. So, t = 20 sec

From equation, H = ut + (½) g t^{2}

As u is 0 (released from rest, assumed), the equation becomes H = (½) g t^{2 }

So,we get H= ½ . 10. 20^{2} = 2000 m……..(answer of i)

The time when echo is heard after the pebble is dropped =

time required by the pebble to touch the water surface after being released + the time required by the sound wave generated to traverse the height of 2000 m to reach the person

= 20 sec + 2000/330 sec

= (20 + 6.06) sec= 26.06 seconds…………..(answer of ii)