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(B) According to Torricelli's law,the velocity of efflux $v$ is given by $v = \sqrt{2gy}$.The time taken for the liquid to reach the ground from a hei

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$x$ is maximum for $y = \frac{H}{2}$.

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(B) According to Torricelli's law,the velocity of efflux $v$ is given by $v = \sqrt{2gy}$.The time taken for the liquid to reach the ground from a height $(H-y)$ is $t = \sqrt{\frac{2(H-y)}{g}}$.The horizontal range $x$ is given by $x = v \cdot t = \sqrt{2gy} \cdot \sqrt{\frac{2(H-y)}{g}} = 2\sqrt{y(H-y)}$.To find the maximum range,we differentiate $x^2 = 4(Hy - y^2)$ with respect to $y$ and set it to zero:$\frac{d}{dy}(4Hy - 4y^2) = 4H - 8y = 0 \implies y = \frac{H}{2}$.Substituting $y = \frac{H}{2}$ into the expression for $x$,we get $x_{max} = 2\sqrt{\frac{H}{2}(H - \frac{H}{2})} = 2\sqrt{\frac{H^2}{4}} = H$.Therefore,$x$ is maximum at $y = \frac{H}{2}$ and the maximum value is $H$.Thus,option $B$ is the correct answer.

$A$ cylindrical vessel is filled with a liquid up to height $H$. $A$ small hole is made in the vessel at a distance $y$ below the liquid surface as shown in the figure. The liquid emerging from the hole strikes the ground at a distance $x$.

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