A vertical cylinder is filled with liquid. A small hole is made in the wall of the cylinder at a depth $H$ below the free surface of the liquid. The force exerted on the cylinder by the liquid flowing out of the hole initially will be proportional to :

\sqrt{H}

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H

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H^{3 / 2}

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H^{2}

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Solution

The correct option is B $H$
By using Bernoulli’s principle

The velocity of the liquid coming out is given as,

$v = \sqrt{2 g h}$

The mass of the liquid will be

$m = ρ A V$

The change in momentum will be

$P = m v$

The force will be proportional of $v^{2}$ which in turn is proportional to $h$

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