A tank is filled up to a height h with a liqu | vedclass

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(D) The velocity of efflux at depth $y$ is $v = \sqrt{2gy}$.The height of the hole from the ground is $H_{ground} = h - y + h = 2h - y$.The time taken

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$A$ and $C$ both

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(D) The velocity of efflux at depth $y$ is $v = \sqrt{2gy}$.The height of the hole from the ground is $H_{ground} = h - y + h = 2h - y$.The time taken by the liquid to reach the ground is $t = \sqrt{\frac{2H_{ground}}{g}} = \sqrt{\frac{2(2h - y)}{g}}$.The horizontal range $x$ is given by $x = v 	imes t = \sqrt{2gy} 	imes \sqrt{\frac{2(2h - y)}{g}} = 2\sqrt{y(2h - y)} = 2\sqrt{2hy - y^2}$.To maximize the range $x$,we differentiate $x^2 = 4(2hy - y^2)$ with respect to $y$ and set it to zero:$\frac{d}{dy}(8hy - 4y^2) = 8h - 8y = 0 \implies y = h$.Substituting $y = h$ into the range equation:$x_m = 2\sqrt{h(2h - h)} = 2\sqrt{h^2} = 2h$.Thus,both $y = h$ and $x_m = 2h$ are correct.

$A$ tank is filled up to a height $h$ with a liquid and is placed on a platform of height $h$ from the ground. To get the maximum range $x_m$,a small hole is punched at a distance of $y$ from the free surface of the liquid. Then:

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