Question

A light cylindrical vessel is kept on a horizontal surface. Its base area is \[A\] . A hole of cross sectional area \[a\] is made just at its bottom side. The minimum coefficient of friction is necessary for sliding of the vessel due to the impact force of the emerging liquid is:
A. Varying
B. $\dfrac{a}{4}$
C. $\dfrac{{2a}}{4}$
D. None of these

Answer 1

Hint:Before going to the question let us first talk about friction. Friction is a force that slides or attempts to slide between two areas. For instance, friction makes it difficult to push a book around the floor. Friction always works the other way in the direction the object moves or tries to move.

Complete step by step answer:
In fluid dynamics the principle of Bernoulli states that the fluid speed increases at the same time as the static pressure decreases, or that the potential energy of the fluid decreases.The velocity of the emerging liquid from the floor is determined if the liquid's height is \[h\] :
$v = \sqrt {2gH} $ (Easily proved by Bernaulli's theorem)
If the density of the liquid is $\rho $
The liquid mass rate that emerges from the vessel
$\dfrac{{dm}}{{dt}} = a \times \sqrt {2gh} \times \rho \\
\Rightarrow \dfrac{{dm}}{{dt}} = a\rho \sqrt {2gh} \\ $
The Second Law of Motion of Newton stipulates that the rate of dynamic change in an object is proportional to the force applied by unequal force. \[F = ma\] , that is. Where \[F\] is applied, \[m\] is the body's weight, and an is produced by acceleration.
From the second law of Newton, Force is
$F = \dfrac{{d\left( {mv} \right)}}{{dt}} \\
\Rightarrow F = v\dfrac{{dm}}{{dt}} \\
\Rightarrow F = \sqrt {2gh} \times a\rho \times \sqrt {2gh} \\
\Rightarrow F = 2gh \times a\rho \\ $
The frictional force must be the same i.e.
$\mu N = 2gh \times a\rho \\
\Rightarrow \mu \times A \times h \times \rho \times g = 2gh \times a\rho \\
\therefore \mu = \dfrac{{2a}}{A} \\ $
Hence, the correct option is C.

Note:Gases and liquids are most often not able to have a negative absolute pressure or even zero pressure, and thus the equation of Bernoulli clearly ceases to be valid prior to zero pressure. Cavitation occurs in liquids, if the pressure is too low.