Q 11-05-050JEE MainJEE Main 2026 (6 Apr, Shift 1)Medium
A smooth inclined plane ends in a vertical circular loop, as shown in the figure. A small body is released from height $h$ as shown. If the body exerts a force of three times its weight on the plane at the highest point of circle then the height $h = \alpha R$. The value of $\alpha$ is ______
Answer: (B) $4$
At the top of the loop the normal force is $3mg$, and it acts downward together with gravity:
$$3mg + mg = \frac{mv^2}{R} \;\Rightarrow\; v^2 = 4gR$$
Energy conservation from the release point to the top (height $2R$):
$$mg(h - 2R) = \frac{1}{2}m(4gR) \;\Rightarrow\; h = 4R$$
So $\alpha = 4$.
Solution by Sreeraj P, M.Sc Physics