First slide

Fluid dynamics

Question

A vessel of area of cross-section $A$ has liquid to a height $H$ . There is a hole at the bottom of vessel having area of cross-section $a$ . The time taken to decrease the level from $H_{1}$ to $H_{2}$ will be:

Moderate

A

$\frac{A}{a} \sqrt{\frac{2}{g}} (\sqrt{H_{1}} - \sqrt{H_{2}})$
B

$\sqrt{2 g h}$
C

$\sqrt{2 g h (H_{1} - H_{2})}$
D

$\frac{A}{a} \sqrt{\frac{g}{2}} (\sqrt{H_{1}} - \sqrt{H_{2}})$

Solution

The average velocity of efflux
$v = \frac{\sqrt{2 g H_{1}} + \sqrt{2 g H_{2}}}{2}$

Let $t$ be the time taken to empty the tank from level $H_{1}$ to $H_{2}$ . Then,

$\frac{\sqrt{2 g H_{1}} + \sqrt{2 g H_{2}}}{2} \times a \times t = A [H_{1} - H_{2}]$

$\sqrt{\frac{g}{2}} [\sqrt{H_{1}} + \sqrt{H_{2}}] a t$

$= A [\sqrt{H_{1}} + \sqrt{H_{2}}] [\sqrt{H_{1}} - \sqrt{H_{2}}]$

or $t = \frac{A}{a} \sqrt{\frac{2}{g}} \times (\sqrt{H_{1}} - \sqrt{H_{2}})$

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