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 with which the liquid comes out of the vessel 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 $\sqrt{2(gh+\frac{mg}{\rho A})}$

Applying Bernoulli theorem across 1 and 2,

$\frac{2mg}{A\rho }+2\rho gh=\frac{1}{2}\rho v^2$

$v_2=\frac{2mg}{A\rho }+2gh$

$v^2=2g(h+\frac{m}{\rho A})$

$v=\sqrt{2(gh+\frac{mg}{\rho A})}$