Water is filled in a container of uniform area of cross section 'A'. A hole of cross sectional area $^{'} a^{'} (a << A)$ is made in the container at a height of 20 m above the base. Water streams out and hits a small block on the surface at some distance from the container. The block is moved on the surface in such a way that the stream always hits the block. The initial velocity of the block (in m/s) is
$(Given \frac{a}{A} = \frac{1}{20} . Take g = 10 {ms}^{- 2})$ .

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Solution

Velocity of efflux $v = \sqrt{2 g y}$
Range $x = \sqrt{2 g y} \times \sqrt{\frac{2 h}{g}}$
The velocity of the block must be $(\frac{d x}{d t})$ .
$V_{b} = \frac{d x}{d t} = \sqrt{\frac{2 h}{g}} \times \sqrt{2 g} \times \frac{1}{2 \sqrt{y}} \frac{d y}{d t}$
$V_{b} = \frac{\sqrt{h}}{\sqrt{y}} . \frac{d y}{d t} \dots \dots (i)$
Using equation of continuity,
$\frac{A d y}{d t} = a \sqrt{2 g y} \dots \dots (i i)$
From equation (i) and (ii),
$V_{b} = \sqrt{\frac{h}{y}} \times \frac{a}{A} \sqrt{2 g y}$
$= \sqrt{2 g h} \times \frac{a}{A} = 20 \times \frac{1}{20} = 1 {ms}^{- 1}$

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