Question

A cylindrical vessel filled with water up to a height of $2$m stands on a horizontal plane. The side wall of the vessel is a plugged circular hole touching the bottom. Find the minimum diameter of the hole so that the vessel begins to move on the floor if the plug is removed. The coefficient of friction between the bottom of the vessel and the plane is $0.4$ and the total mass of water plus vessel is $100$ kg.

Answer 1

The velocity of efflux through the hole,$v=\surd \left(2gh\right)$

Let rho be the density and A be the area of cross section of the hole

Therefore Rate of momentum =$(Rho\times A\times v)\times \left(v\right)$

=$2ghA\rho [usingv=\surd (2gh\left)\right]$

Acc. To Newton 2nd Law of Motion,

Force due to vel. Of efflux =$2ghA\rho$

The vessel will move if Force on vessel=Force of friction

$2ghA\rho =\left(\mu \right)MgA$

=$\left(\mu \right)\frac{M}{2hrho}$

A=$\frac{1}{100}$

Also, $A=\pi r^2=\frac{\pi }{D^2}\times 4$

Therefore $D=\surd \frac{4A}{\pi }$

=>$D=0.11metre$