Question

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

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

Answer 1

The correct option is C $\sqrt{\frac{2h}{g}}+\frac{h}{v}$

Time taken for the sound to reach to the top = Time taken for the stone to reach the ground + Time taken for the sound to reach the top.
$h=ut+\frac{1}{2}a{t}^2$ (u is zero since the stone is freely falling)
$h=\frac{1}{2}a{t}^2$
$t_g=\sqrt{\frac{2h}{g}}$
Time taken for the sound to reach the top is
$t_c=\frac{h}{v}$
Total time taken $=\sqrt{\frac{2h}{g}}+\frac{h}{v}$