Question

A Cylindrical vessel filled with water upto height of h stands on a horizontal plane. The side wall of the vessel has a plugged circular hole touching the bottom. The coefficient of friction between the bottom of vessel and plane is $\mu$ and total mass of water plus vessel is m. What should be minimum diameter of hole so that the vessel begins to move on the floor if plug is removed (here density of water is $\rho$):

• A. $\sqrt{\frac{2\mu m}{\pi \rho h}}$
• B. $\sqrt{\frac{\mu m}{2\pi \rho h}}$
• C. $\sqrt{\frac{\mu m}{\rho h}}$
• D. none

Answer 1

The correct option is A $\sqrt{\frac{2\mu m}{\pi \rho h}}$

From Toricelli's theorem we get the velocity of the water from the hole as $v=\sqrt{2gh}$
Impulse of force acting from the side of the vessel on out flowing water is
$F\Delta t=\Delta \left(mv\right)=\left(\Delta m\right)v$(here $\Delta m=\rho Av\Delta t$)
or
$F=\rho A{v}^2=\rho A\left(2gh\right)$
Now the vessel to move this force $F$ should be greater than the frictional force. Thus
$2\rho ghA>\mu mg$
or
$A>\frac{\mu m}{2\rho h}$
or
$r>\sqrt{\frac{\mu m}{\pi 2\rho h}}$
or
$d>\sqrt{\frac{2\mu m}{\pi \rho h}}$