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 :
$$T = \sqrt {\left( {\frac{{2h}}{g}} \right)} + \frac{h}{v}$$
Solution :
Time taken by the stone to reach the water level $${t_1} = \sqrt {\frac{{2h}}{g}} $$
Time taken by sound to come to the mouth of the well, $${t_2} = \frac{h}{v}$$
$$\therefore $$ Total time $${t_1} + {t_2} = \sqrt {\frac{{2h}}{g}} + \frac{h}{v}$$