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 velocity of sound, the time $T$ after which the splash is heard is given by

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

Answer 1

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

Time taken by the stone to reach the water level,
By second kinematic equation,
$S=ut+\frac{1}{2}g{t}^2\Rightarrow t_1=\sqrt{\frac{2h}{g}}-------\left(i\right)$
Time taken by sound to come to the mouth of the well,
$Time=\frac{Distance}{Speed}\Rightarrow t_2=\frac{h}{v}-------\left(ii\right)$
$\therefore$ Total time $t_1+t_2=\sqrt{\frac{2h}{g}}+\frac{h}{v}$