Question

A stone is dropped from a tower of height H metre on the upper surface of a well having water at depth d metre. If g is the acceleration due to gravity, then the time when the sound of splashing will be heard from the time of dropping the stone is [Take speed of sound in air = 330 m/s]

• A. $[\sqrt{\frac{H+d}{2g}}+\left(\frac{H+d}{330}\right)]s$
• B. $[\sqrt{\frac{H+d}{g}}+\left(\frac{H+d}{330}\right)]s$
• C. $[\sqrt{\frac{2(H+d)}{g}}+\left(\frac{H+d}{330}\right)]s$
• D. $[\sqrt{\frac{2H}{g}}+\left(\frac{H}{330}\right)]s$

Answer 1

The correct option is C $[\sqrt{\frac{2(H+d)}{g}}+\left(\frac{H+d}{330}\right)]s$

Let $t_1$ be the time taken by the stone to cover the distance $d+H$ and $t_2$ be the time taken by the sound to come back to the dropping point by covering the distance $d+H.$

$\therefore$ Time when the splashed sound is heard, $T=t_1+t_2$

For downward motion of stone:

Initial velocity, u = 0 m/s

Acceleration, a = g

Distance covered, s = d + H

Using second equation of motion, $s=ut+\frac{1}{2}a{t}^2\Rightarrow d+H=0\times t+\frac{1}{2}g(t_1{)}^2\Rightarrow t_1=\sqrt{\frac{2(d+H)}{g}}....(i)$

Sound travels a distance $(s=d+H)$ to reach the dropping point of stone. Speed of sound, $v=330m/s$

Time taken by sound,$t_2=\frac{s}{v}=\frac{d+H}{330}...\left(ii\right)\Rightarrow T=t_1+t_2=[\frac{\sqrt{2(H+d)}}{g}+\left(\frac{H+d}{330}\right)]s$

Hence, the correct answer is option (c).