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 into the container at a constant rate of $Q\text{m}{}^3\text{/s}.$ The maximum level of liquid in the container will be -

• A. $\frac{Q^2}{2g{A}_1{A}_2}$
• B. $\frac{Q^2}{2g{A}_1^2}$
• C. $\frac{Q}{2g{A}_1{A}_2}$
• D. $\frac{Q^2}{2g{A}_2^2}$

Answer 1

The correct option is B $\frac{Q^2}{2g{A}_1^2}$

Given,
Rate of inflow, $Q_{in}=Q\text{m}{}^3\text{/s}$ ........(1)
According to question -


We know that, level in the container is maximum when rate of inflow $=$ rate of outflow.
Rate of outflow, $Q_{out}=Av$ ........(2)
where $A$ is the area of the hole and $v$ is the velocity of efflux.
From (1) and (2) -
$A_{1}v=Q$ $[A=A_1,$ given $]$
$\Rightarrow A_1\sqrt{2gh}=Q$
$\Rightarrow h=\frac{Q^2}{2g{A}_1^2}$