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 $v_{s o u n d} = 340 m s^{- 1}$ .

$180 m$

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$100 m$

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$90 m$

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$220 m$

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Solution

The correct option is A
$180 m$

Given,
initial velocity, $u = 0$
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, $s = u t + \frac{1}{2} a t^{2}$
$D = \frac{1}{2} \times 10 \times T_{1}^{2}$

$T_{1}$ = $\sqrt{2 \times \frac{D}{10}}$ = $\sqrt{\frac{D}{5}}$

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

Total time, $T_{1} + T_{2}$ = $\sqrt{\frac{D}{5}}$ + $\frac{D}{340} = 6.53$

On solving we get, $D = 180 m$ .

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Tinku dropped a stone into a very deep well. He hears the sound 6.53 s after he had dropped the stone. Find the depth of the well. Take acceleration due to gravity, g = 10m/s² and Vsound as 340 m/s

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Q. 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}

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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 ms−2 and vsound=340 ms−1.

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 $v_{s o u n d} = 340 m s^{- 1}$ .