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A container has a small hole at its bottom. Area of cross-section of the hole is A₁ and that of the container is A₂. Liquid is poured in the container at a constant rate Q m³s^-1. The maximum level of liquid in the container will be

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$\frac{Q}{2 g A_{1} A_{2}}$

$\frac{Q^{2}}{2 g A_{1} A_{2}}$

$\frac{Q^{2}}{2 g A_{2}^{2}}$

$\frac{Q^{2}}{2 g {A_{1}}^{2}}$

answer is B.

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Detailed Solution

Level in the container will become maximum, when rate of inflow = rate of outflow

$Q = A_{1} v = A_{1} \sqrt{2 g h_{max}}$

$∴ h_{max} = \frac{Q^{2}}{2 g A_{1}^{2}}$

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