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(B) Let $y$ be the depth of a hole below the water surface. The velocity of efflux is $v = \sqrt{2gy}$.The height of the hole from the ground is $H_{h

Vedclass · Accepted Answer

$2$

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(B) Let $y$ be the depth of a hole below the water surface. The velocity of efflux is $v = \sqrt{2gy}$.The height of the hole from the ground is $H_{hole} = h - y + h/2 = \frac{3h}{2} - y$.The time taken for water to reach the plane $PQ$ is $t = \sqrt{\frac{2 H_{hole}}{g}} = \sqrt{\frac{2(3h/2 - y)}{g}} = \sqrt{\frac{3h - 2y}{g}}$.The horizontal range $x$ is given by $x = v \cdot t = \sqrt{2gy} \cdot \sqrt{\frac{3h - 2y}{g}} = \sqrt{2y(3h - 2y)} = \sqrt{6hy - 4y^2}$.To maximize $x$,we differentiate with respect to $y$ and set it to zero: $\frac{dx}{dy} = \frac{1}{2\sqrt{6hy - 4y^2}} \cdot (6h - 8y) = 0$.This gives $6h - 8y = 0$,or $y = \frac{3h}{4}$.The depth $y$ is measured from the top. The heights of the holes from the base are $0, h/4, h/2, 3h/4$. Thus,the depths $y$ from the top are $h, 3h/4, h/2, h/4$ for holes $1, 2, 3, 4$ respectively.The range is maximum when $y = 3h/4$,which corresponds to hole number $2$.

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