Question

Tinku dropped a stone into a very deep well. He hears the sound 6.53 s after he dropped the stone. Find the depth of the well. Take acceleration due to gravity, g =10 $m{s}^{-2}$ and $v_{sound}$ = 340 $m{s}^{-1}$.

• A. 180 m
• B. 100 m
• C. 90 m
• D. 220 m

Answer 1

The correct option is A

180 m

Initial velocity, u = 0, D = depth of well

$T_1$ = time taken by the stone to reach depth D when it is thrown

$T_2$ = time taken by sound to reach the surface

Using second equation of motion, S = $ut+\frac{1}{2}g{t}^2$

$D=\frac{1}{2}\times 10\times T_1^2$

$T_1$ = $\sqrt{2\times \frac{D}{10}}$

Now, $T_2$ = $\frac{D}{340}$(Since $\text{time}=\frac{\text{Distance}}{\text{Speed}}$)

As per the question, $T_1$ + $T_2$ = 6.53

$\sqrt{\frac{D}{5}}$ + $\frac{D}{340}$ = 6.53

Solving we get, D = 180 m.