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A horizontal pipe is running full of water. At a point 'A’ it tapers from $400 c m^{2}$ cross - sectional area to $200 c m^{2}$ at other point 'B' . If the pressure difference between A and B is $30 c m$ of water column the rate of volume flow of water through the pipe is $(g = 10 m s^{- 2})$

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$5 \sqrt{2} lit / sec$

$40 \sqrt{2} l i t / \sec$

$30 l i t / \sec$

$15 l i t / \sec$

answer is A.

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Detailed Solution

According to Bernoulli’s theorem

$\frac{p_{1} - p_{2}}{d} = \frac{1}{2} (V_{2}^{2} - V_{1}^{2}) \Rightarrow \frac{30 \times 10^{- 2} \times 1000 \times 10}{1000} = \frac{1}{2} (V_{2}^{2} - V_{1}^{2}) \Rightarrow V_{2}^{2} - V_{1}^{2} = 6$

$\Rightarrow \frac{V_{1}}{V_{2}} = \frac{A_{2}}{A_{1}} = \frac{200}{400} = \frac{1}{2}$

$\Rightarrow 4 V_{1}^{2} - V_{1}^{2} = 6 V_{1} = \sqrt{2}$

$∴ V = a_{1} v_{1}$ $= 400 \times \sqrt{2} \times 10^{- 4} \times 10^{3}$ $= 40 \sqrt{2} l i t / \sec$

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