JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Laws of Motion

The mass presses against the inner wall, so the wall's normal reaction points towards the axis and provides the entire centripetal force:

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

Vertically, static friction from the wall supports the weight:

$f=mg$, with $f\le \mu N$

The mass does not slide down provided $\mu N\ge mg$, i.e. $\mu m{\omega }^{2}R\ge mg$, giving $\omega \ge \sqrt{\frac{g}{\mu R}}$.

At the minimum speed the friction is limiting, so equality holds:

$\mu =\frac{g}{{\omega }^{2}R}=\frac{10}{(5{)}^2\times 4}=\frac{10}{100}=0.1$

Note that the mass 0.5 kg cancels out — $\mu$ does not depend on it.

Hence option (A).