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: (Where $A_1 << A_2$)

• 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

Answer: (B)

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

Answer 2

The maximum level of liquid in the container is reached when the rate of liquid flowing into the container equals the rate of liquid flowing out through the hole.

Inflow rate = $Q$.

Outflow rate = Area of hole $\times$ velocity of efflux.

Using Torricelli's theorem, the velocity of efflux from a hole at depth $h$ is $v = \sqrt{2gh}$, assuming the area of the container is much larger than the area of the hole.

Outflow rate = $A_1 \sqrt{2gh}$.

At the maximum height $h_{max}$, $Q = A_1 \sqrt{2gh_{max}}$.

Squaring both sides, $Q^2 = A_1^2 (2gh_{max})$.

Solving for $h_{max}$, we get $h_{max} = \frac{Q^2}{2g A_1^2}$.