A container has a small hole at its bottom. Area of cross-section of the hole is `A_(1)` and that of the container is `A_(2)`. Liquid is poured in the container at a constant rate ` Q m^(3)s^(-1)`. The maximum level of liquid in the container will be

Options

A. `(Q^2)/(2gA_1A_2)`
B. `(Q^2)/(2gA_1^2)`
C. `(Q)/(2gA_1A_2)`
D. `(Q^2)/(2gA_2^2)`

Answer

B

Explanation

Level in the container will become maximum when rate of in flow = rate of out flow
`Q=A_1v=A_1sqrt(2gh_"max"))`
`therefore h_("max") = (Q^2)/(2ghA_1^2)`