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

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$T = \frac{2 h}{v}$

$T = \sqrt{(\frac{h}{2 g})} + \frac{2 h}{v}$

$T = \sqrt{(\frac{2 h}{g})} + \frac{h}{v}$

$T = \sqrt{(\frac{2 h}{v})} + \frac{h}{g}$

answer is B.

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Detailed Solution

Time taken by the stone to reach the water level

$t_{1} = \sqrt{\frac{2 h}{g}}$

Time taken by sound to come to the mouth of the well, $t_{2} = \frac{h}{v}$

∴ Total time $t_{1} + t_{2} = \sqrt{(\frac{2 h}{g})} + \frac{h}{v}$

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