Question

A light cylindrical vessel is kept on a horizontal surface. Area of base is $A$. A hole of cross-sectional area ${}^{\prime }{a}^{\prime }$ is made just at its bottom side. The minimum coefficient of friction necessary to prevent sliding the vessel due to the impact force of the emerging liquid is $(a<<A)$ :

• A. $\frac{A}{2a}$
• B. None of these
• C. $\frac{2a}{A}$
• D. $\frac{a}{A}$

Answer 1

The correct option is C $\frac{2a}{A}$

Exit speed of liquid is given by

$v^2=2gh$

Where $h$ is the height of liquid

So, impact force of emerging liquid,

$F=\rho a{v}^2=2\rho agh$

For no sliding condition,

$f\ge F$

Where, $f$ is the force of friction.
$f=\mu mg=\mu \left(\rho Ah\right)g$

$\Rightarrow \mu \rho Ahg\ge 2\rho agh$

$\Rightarrow \mu \ge \frac{2a}{A}$

So the minimum coefficient of friction for the vessel is,

$\begin{array}{c}\therefore \mu =\frac{2a}{A}\end{array}$

Hence, option $\left(\text{C}\right)$ is correct.