JEE Main 2026 — Laws of Motion Question with Solution

The normal force $N$ provides the necessary centripetal force for the mass to move in a circle:

$N=m{\omega }^{2}R$

For the mass to remain stuck to the wall without falling, the upward frictional force must balance the downward gravitational force:

$f\ge mg$

Since the maximum static friction is $f_{max}=\mu N$, we have:

$\mu N\ge mg$

$\mu (m{\omega }^{2}R)\ge mg$

$\mu \ge \frac{g}{{\omega }^{2}R}$

Substituting the given values $g=10$ m/s${}^2$, $\omega =5$ rad/s, and $R=4$ m:

$\mu \ge \frac{10}{5^2\times 4}$

$\mu \ge \frac{10}{100}$

$\mu \ge 0.1$

The minimum coefficient of friction is $0.1$.

Answer: $0.1$