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

Answer 1

Hint:The total time taken for the sound of splash to be heard from the moment of its release is given. The depth of the well, h, could be expressed in terms of the time it takes for a stone to fall to the bottom using Newton's equation of motion. Now find the time taken for the sound of impact to reach the top with given velocity of sound in air and h.

Formula used:
Newton’s equation of motion,
$s = ut + \dfrac{1}{2}g{t^2}$
Here, $s$ is the displacement of the particle, $u$ is the initial velocity, $g$ is the acceleration of gravity and $t$ is the time period.

Complete step by step solution:
In the question, we are given a stone that is dropped into the well. The sound of the splash of the stone when it hits the water is heard after $T$ its release from top. We are given the acceleration due to gravity and velocity of sound in air $g$ and $v$ respectively. Height of the well is $h$ meters.

Now since the stone is dropped the initial velocity will be zero ($\therefore u = 0$). So, time taken ${t_1}$ by the stone to reach the bottom of the well is given by newton's equation of motion.
$ \Rightarrow h = \dfrac{1}{2}g{t_1}^2$
$ \Rightarrow {t_1} = \sqrt {\dfrac{{2h}}{g}} $
Now the time taken ${t_2}$ by the stone splash sound to come at the mouth of well is given by,
${t_2} = \dfrac{h}{v}$
where $v$ is the speed of sound
So, the total time taken will be
$T = {t_1} + {t_2} = \sqrt {\dfrac{{2h}}{g}} + \sqrt {\dfrac{h}{v}} \\ $
Hence option C is correct.

Note: All we need to know in this case is that the time it takes for the sound of splash to go to the top will be equal to the sum of the times it takes for the stone to reach the bottom and the sound to do so. Knowing this and making the straightforward substitutions required will result in the solution. As we read the question, always try to picture what is really happening.