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An orifice is present on the vertical wall of a water tank at a depth h below the free surface of water. When h = 4m, rate of flow of water through the orifice is 0.25 $\frac{m^{3}}{s}$ . What will be the rate of flow of water when h = 9m?

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0.825 $m^{3} / s$

0.625 $m^{3} / s$

0.425 $m^{3} / s$

0.375 $m^{3} / s$

answer is D.

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Detailed Solution

If ‘a’ be the cross-sectional area of the orifice, then rate of flow $Q = a \sqrt{2 g h}$

$∴ Q \propto \sqrt{h} \Rightarrow \frac{Q^{1}}{Q} = \sqrt{\frac{h^{1}}{h}} \Rightarrow \frac{Q^{1}}{0.25} = \sqrt{\frac{9}{4}} \Rightarrow Q^{1} = 0.375 \frac{m^{3}}{s}$

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