A tank is filled upto a height $h$ with a liquid and is placed on a platform of height h from the ground. To get maximum range ${x_m}$ a small hole is punched at a distance of $y$ from the free surface of the liquid. Then
Diffcult
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(d) Velocity of liquid through orifice, $v = \sqrt {2gy} $
and time taken by liquid to reach the ground
$t = \sqrt {\frac{{2(h + h - y)}}{g}} = \sqrt {\frac{{2(2h - y)}}{g}} $
$\therefore$ Horizontal distance covered by liquid
$x = v.t. = \sqrt {2gy} \times \sqrt {\frac{{2(2h - y)}}{g}} = \sqrt {4y(2h - y)} $
==> ${x^2} = 4y(2h - y)$
==> $\frac{{d{{(x)}^2}}}{{dy}} = 8h - 8y$
for $x$ to be maximum, $\frac{d}{{dy}}({x^2}) = 0$
$\therefore 8h - 8y = 0$ or $h = y$
So ${x_m} = \sqrt {4h(2h - h)} = 2h$
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