A stone is dropped into a well and the sound of impact of stone with the water is heard after $2.056 s$ of the release of stone from the top. If the acceleration due to gravity is $980 cm s^{- 2}$ and the velocity of sound in air is $350 {ms}^{- 1}$ , then the depth of the well is

7 m

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19.6 m

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700 m

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1960 m

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Solution

The correct option is C 19.6 m
Time taken by the stone to reach the water surface
$t_{1} = \sqrt{\frac{2 h}{g}}$ , where h is the height of the well.
Time taken by the sound to reach the observer
$t_{2} = \frac{h}{350}$
Total time
$t = t_{1} + t_{2} = \sqrt{\frac{2 h}{g}} + \frac{h}{350} = \sqrt{\frac{h}{4.9}} + \frac{h}{350} = 2.056$
Solving the above Eqn. we get
$h = 19.6 m$

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A stone is dropped into a well and the sound of impact of stone with the water is heard after 2.056 s of the release of stone from the top. If the acceleration due to gravity is 980 cms−2 and the velocity of sound in air is 350 ms−1, then the depth of the well is

The correct option is C 19.6 mTime taken by the stone to reach the water surface t1=√2hg, where h is the height of the well.Time taken by the sound to reach the observer t2=h350Total time t=t1+t2=√2hg+h350=√h4.9+h350=2.056Solving the above Eqn. we get h=19.6 m

A stone is dropped into a well and the sound of impact of stone with the water is heard after $2.056 s$ of the release of stone from the top. If the acceleration due to gravity is $980 cm s^{- 2}$ and the velocity of sound in air is $350 {ms}^{- 1}$ , then the depth of the well is