Question

A cylindrical vessel contains a liquid of density $\rho$ up to a height $H$. The liquid is closed by a piston of mass $m$ and area of cross-section $A$. There is a small hole at the bottom of the vessel. The speed $v$ with which the liquid comes out of the hole is

• A. $\sqrt{3(gH+\frac{mg}{\rho A})}$
• B. $\sqrt{2(gH+\frac{mg}{\rho A})}$
• C. $\sqrt{5(gH+\frac{mg}{\rho A})}$
• D. $\sqrt{6(gH+\frac{mg}{\rho A})}$

Answer 1

The correct option is B $\sqrt{2(gH+\frac{mg}{\rho A})}$

Total pressure at position $1$
$p_1=p_0+\frac{mg}{A}$

As area of cross-section of container is very large compared to area of cross-section of hole.
So, $v_1\approx 0$
[from equation of continuity]
Applying Bernoulli's theorem at both positions $1$ and $2$.
$p_1+\frac{1}{2}\rho v_1^2+\rho g{h}_1=p_2+\frac{1}{2}\rho v_2^2+\rho g{h}_2$
$\Rightarrow p_0+\frac{mg}{A}+0+\rho gH=p_0+\frac{1}{2}\rho v^2+0$
$\Rightarrow p_0+\frac{mg}{A}+\rho gH=p_0+\frac{1}{2}\rho v^2$
$\Rightarrow v^2=2(gH+\frac{mg}{\rho A})$
$\Rightarrow v=\sqrt{2(gH+\frac{mg}{\rho A})}$