Question

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
A. `(Q^(2))/(2gA_(1)A_(2))`
B. `(Q^(2))/(2gA_(1)^(2))`
C. `(Q)/(2g A_(1)A_(2))`
D. `(Q^(2))/(2g A_(2_^(2))`

Answer 1

Correct Answer - B
Level in the container will become maximum when rate of inflow = rate of outflow.
`Q=A_(1)v=A_(1)sqrt(2gh_("max"))`
`therefore" "h_("max")=Q^(2)/(2gA_(1)^(2))`