Question

A cylindrical vessel contains a liquid of density $\rho$ upto 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. $\sqrt{2gh}$
• B. $\sqrt{2(gh+\frac{mg}{\rho A})}$
• C. $\sqrt{2(gh+\frac{mg}{A})}$
• D. $\sqrt{2gh+\frac{mg}{A}}$

Answer 1

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


Let atmospheric pressure be $P_o$.
Pressure at point $1=P_o+\frac{mg}{A}+\rho gh$
Pressure at point $2=$ Atmospheric pressure $=P_o$
Applying Bernoulli's theorem at point $1$ and $2$, we get
$(P_o+\rho gh)+\frac{mg}{A}+\rho g{h}_1=P_o+\frac{1}{2}\rho v^2+\rho g{h}_2$
(Since $h_1=h_2$)
$\Rightarrow v=\sqrt{2(gh+\frac{mg}{\rho A})}\text{m/s}$