Question

29. A light cylindrical vessel is kept on a horizontal
surface. Its base area is A. A hole of cross
sectional area
is made on side wall just at its
bottom. Minimum coefficient of friction such that it
does not slide due to impact force of emerging
liquid is

Answer 1

Answer:

The minimum coefficient of friction is $\frac{2a}{A}$

Explanation:

If the height of the liquid in the vessel is h then the velocity of the emerging liquid from the bottom is given by

$v=\sqrt{2gh}$    (Easily proved by Bernaulli's theorem)

if the density of the liquid is $\rho$

The rate of mass of liquid emerging out of the vessel

$\frac{dm}{dt}=a\times \sqrt{2gh}\times \rho$

$\implies \frac{dm}{dt}=a\rho\sqrt{2gh}$

From the second law of Newton, Force is

$\boxed {F=\frac{d(mv)}{dt}}$

$\implies F=v\frac{dm}{dt}$

$\implies F=\sqrt{2gh}\times a\rho\times \sqrt{2gh}$

$\implies F=2gh\times a\rho$

This must be equal to the frictional force

i.e.

$\mu N=2gh\times a\rho$

$\implies \mu\times A\times h\times \rho\times g=2gh\times a\rho$

$\implies \mu=\frac{2a}{A}$

Hope this helps.