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A water barrel having water up to a depth ‘d’ is placed on a table of height ‘h’. A small hole is made on the wall of the barrel at it’s bottom. If the stream of water coming out of the hole falls on the ground at a horizontal distance ‘R’ from the barrel, then the value of ‘d’ is

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$\frac{4 h}{h^{2}}$

4 hR²

$\frac{R^{2}}{4 h}$

$\frac{h}{4 R^{2}}$

answer is C.

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Detailed Solution

$R = u \times t = \sqrt{2 g d} \times \sqrt{\frac{2 h}{g}}$

$R^{2} = 2 g d \times \frac{2 h}{g} = 4 h d$

$\begin{matrix} d = \frac{R^{2}}{4 h} \end{matrix}$

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