Q 11-06-095JEE MainJEE Main 2024 (9 Apr, Shift 2)Easy
A circular disc reaches from top to bottom of an inclined plane of length $l$. When it slips down the plane, it takes $t$ s. When it rolls down the plane, it takes $\left(\dfrac\alpha2\right)^{1/2}t$ s, where $\alpha$ is ______.
Numerical value type. Enter your answer.
Answer: 3
Slipping (no friction): $a_1 = g\sin\theta$. Rolling disc: $a_2 = \dfrac{g\sin\theta}{1 + \frac12} = \dfrac23g\sin\theta$.
For the same length from rest, $t \propto \dfrac{1}{\sqrt a}$:
$$t_2 = t\sqrt{\frac{a_1}{a_2}} = \left(\frac32\right)^{1/2}t \;\Rightarrow\; \alpha = 3$$
Solution by Sreeraj P, M.Sc Physics