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Liquid is filled in a container upto a height of H. A small hole is made at the bottom of the tank. Time taken to empty from H to $\frac{H}{3}$ is $t_{0}$ . Find the time taken to empty the $tank$ from $\frac{H}{3}$ to zero.

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None of above

$\frac{A}{a} \sqrt{\frac{2 (H / 3)}{g}}$

$\frac{a}{A} \sqrt{\frac{3 (H / 2)}{g}}$

$\frac{2 A}{a} \sqrt{\frac{3 (H / 5)}{g}}$

answer is B.

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Detailed Solution

At any time t, let the height of water in the tank be h.

At that instant velocity of efflux = $\sqrt{2 g h}$

From principle of continuity , $A (- \frac{d h}{d t}) = a \sqrt{2 g h}$

Integrating from h= $\frac{H}{3} t o h = 0$ , we find t = $\frac{A}{a} \sqrt{\frac{2 H}{3 g}}$

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