Tinku dropped a stone into a well of depth $180 m$ at time $t = 0$ . He heard a sound after time $t = x$ . What is the value of $x$ . Take acceleration due to gravity, $g = 10 m s^{- 2}$ and $v_{s o u n d} = 340 m s^{- 1}$ .

6 s

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6.53 s

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5.53 s

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3 s

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Solution

The correct option is B $6.53 s$
Given,
Initial velocity, $u = 0 m / s$
Depth $D = 180 m$
Acceleration due to gravity, $g = 10 m s^{- 2}$
Speed of sound. $v_{s o u n d} = 340 m s^{- 1}$

$T_{1}$ = time taken by the stone to reach depth D when it is thrown

$T_{2}$ = time taken by sound to reach the surface
From second equation of motion, we have,
$s = u t + \frac{1}{2} a t^{2}$
$D = \frac{1}{2} \times 10 \times T_{1}^{2}$
$T_{1}$ = $\sqrt{2 \times \frac{D}{10}}$ = $\sqrt{\frac{D}{5}}$ = $\sqrt{\frac{180}{5}}$ = $6 s$

Time taken for the sound to travel from bottom of the well to the surface

We know, $t i m e = \frac{d i s t a n c e}{s p e e d}$
$\Rightarrow$ $T_{2}$ = $\frac{180}{340}$ = $0.53 s$

Total time, $T_{1} + T_{2}$ = $6.53 s$

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