Question

A stone is dropped into a well in which the level of water is h, below the top of the well. If v is the velocity of sound, then the time T after which the splash is heard is given by:

1. $\frac{2h}{v}$
2. $\sqrt{ \frac{2h}{g} } + \frac{h}{v}$
3. $\sqrt{ \frac{h}{2g} } + \frac{2h}{v}$
4. $\sqrt{ \frac{2h}{v} } + \frac{h}{g}$

Answer 1

Time taken for thw ball to reach the bottom
= (2h/g)^1/2

Time taken by sound to reach the top
= h/v

Thus total time = (2h/g)^1/2 + h/v

Thus answer is B

Answer 2

Answer:

2. √2h/g + h/v

Explanation:

Let the time taken by the stone to reach to the water level = T

Let T also be the time taken by the sound to come out from the water level to the top of the well.  

Thus, if T1 is the time taken by the stone to reach the water level then,

= h = 1/2 x g x T1²

= T1 = (2h/g)^1/2

Since v is the velocity of the sound, the time taken by velocity to reach the top will be -

T2 = h/v

Total time -

= T = T1 + T2

= (2h/g)^1/2 + h/v

Thus, the time after which the splash is heard is given by: √2h/g + h/v