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A container has a small hole at its bottom. Area of cross-section of the hole is A1 that of the container is A2 . Liquid is poured in the container at a constant rate Q  m3/s. The maximum level of liquid in the container will be

Option: 1

\frac{Q^2}{2 g A_1 A_2}


Option: 2

\frac{Q^2}{2 g A_1^{2}}


Option: 3

\frac{Q}{2 g A_1 A_2}


Option: 4

\frac{Q^2}{2 g A_2^{2}}


Answers (1)

best_answer

level in the container will become maximum when rate of inflow = rate of out flow or

Q=A_1 V=A_1 \sqrt{2 g h_{\text {max }}}

\therefore \quad h_{\text {max }}=\frac{Q^2}{\2g A_1^2}

Posted by

Gaurav

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