Q 11-04-192JEE MainJEE Main 2019 (12 Jan, Shift 2)Medium
A block kept on a rough inclined plane, as shown in the figure, remains at rest up to a maximum force $2$ N down the inclined plane. The maximum external force up the inclined plane that does not move the block is $10$ N. The coefficient of static friction between the block and the plane is: [Take $g = 10\ \text{m/s}^2$]
Answer: (D) $\dfrac{\sqrt3}{2}$
Let $f = \mu mg\cos30^\circ$ be the limiting friction.
Pushed down with $2$ N (friction up the plane): $2 + mg\sin30^\circ = f$.
Pushed up with $10$ N (friction down the plane): $10 = mg\sin30^\circ + f$.
Subtracting: $2mg\sin30^\circ = 8$, so $mg = 8$ N and $f = 6$ N.
$$\mu = \frac{6}{8\cos30^\circ} = \frac{6}{4\sqrt3} = \frac{\sqrt3}{2}$$
Solution by Sreeraj P, M.Sc Physics