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(C) Let the total height of the water level from the plane $PQ$ be $H$. The cylinder has height $h$ and is placed on a block of height $h/2$. Thus, $H

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$3$

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(C) Let the total height of the water level from the plane $PQ$ be $H$. The cylinder has height $h$ and is placed on a block of height $h/2$. Thus, $H = h + h/2 = 3h/2$.Let $y$ be the height of a hole from the base of the cylinder. The depth of this hole from the free water surface is $d = H - y = 3h/2 - y$.The horizontal range $R$ of the water jet is given by $R = 2\sqrt{d \cdot y_{ground}}$, where $y_{ground}$ is the height of the hole from the plane $PQ$. Here, $y_{ground} = y + h/2$.So, $R = 2\sqrt{(3h/2 - y)(y + h/2)}$.To maximize $R$, we maximize the product $(3h/2 - y)(y + h/2)$. Let $f(y) = (3h/2 - y)(y + h/2)$.Taking the derivative with respect to $y$ and setting it to zero: $f'(y) = -(y + h/2) + (3h/2 - y) = 0$.$2y = h$, which gives $y = h/2$.Comparing the given hole heights: Hole $1$ is at $y=0$, Hole $2$ is at $y=h/4$, Hole $3$ is at $y=h/2$, and Hole $4$ is at $y=3h/4$.Thus, Hole $3$ is at the height $y=h/2$ which provides the maximum range.

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