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 speed of sound in the air, the time $T$ after which the splash is heard is given by

• A. $T=\frac{2h}{v}$
• B. $T=\sqrt{\frac{2h}{g}}+\frac{h}{v}$
• C. $T=\sqrt{\frac{2h}{g}}+\frac{h}{2v}$
• D. $T=\sqrt{\frac{h}{2g}}+\frac{2h}{v}$

Answer 1

The correct option is B $T=\sqrt{\frac{2h}{g}}+\frac{h}{v}$

Total time taken to hear the sound of splash = Time taken by stone to reach the water surface inside the well + Time taken by sound of splash to reach the ear through air.

Let $t_1$ be time taken by the stone to reach the water surface
$h=u{t}_1+\frac{1}{2}g{t}_1^2$
As stone is dropped, so, $u=0$
$\therefore h=\frac{1}{2}g{t}_1^2$
$\Rightarrow t_1=\sqrt{\frac{2h}{g}}.......\left(i\right)$

Also, time taken $t_2$ by sound of splash to reach the ear through air is
$\Rightarrow t_2=\frac{h}{v}..........\left(ii\right)$
$\therefore$ Total time taken to hear the sound of splash $T=t_1+t_2=\sqrt{\frac{2h}{g}}+\frac{h}{v}$