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(D) Let the depth of water in the cylinder be $x$.According to Torricelli's law,the velocity of efflux $(v)$ is given by $v = \sqrt{2 g x}$.The water

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$R^2 / 4 H$

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(D) Let the depth of water in the cylinder be $x$.According to Torricelli's law,the velocity of efflux $(v)$ is given by $v = \sqrt{2 g x}$.The water stream acts as a projectile launched horizontally from a height $H$.The time taken $(t)$ by the water to reach the ground is given by the equation of motion $H = \frac{1}{2} g t^2$,which gives $t = \sqrt{\frac{2 H}{g}}$.The horizontal range $(R)$ is the product of the horizontal velocity and the time of flight:$R = v 	imes t$$R = \sqrt{2 g x} 	imes \sqrt{\frac{2 H}{g}}$$R = \sqrt{2 g x \cdot \frac{2 H}{g}}$$R = \sqrt{4 x H}$Squaring both sides,we get $R^2 = 4 x H$.Therefore,the depth of water $x = \frac{R^2}{4 H}$.

$A$ cylinder containing water stands on a table of height $H$. $A$ small hole is punched in the side of the cylinder at its base. The stream of water strikes the ground at a horizontal distance $R$ from the table. Then the depth of water in the cylinder is ............

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