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Water stands at a height H in a large tank whose sides are vertical. A hole is made in one of the walls of the tank at a depth h below the surface of water. The distance R from the foot of the wall where the emerging stream of water strikes the floor is

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$2 \sqrt{h (H - h)}$

$2 \sqrt{h H}$

$\sqrt{h (H - h)}$

$\sqrt{h H}$

answer is C.

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Detailed Solution

If v is the speed with which water flows out of the hole, then we have according to Torricelli’s theorem.

$v = \sqrt{2 g h}$

Further $H - h = \frac{1}{2} g t^{2}$

$\Rightarrow$ $t = \sqrt{\frac{2 (H - h)}{g}}$

$\Rightarrow$ $R = v t = 2 \sqrt{h (H - h)}$

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