Q 11-06-089JEE MainJEE Main 2024 (29 Jan, Shift 1)Easy
A cylinder is rolling down on an inclined plane of inclination $60^\circ$. Its acceleration during rolling down will be $\dfrac{x}{\sqrt3}\ \text{m s}^{-2}$, where $x =$ ______ (use $g = 10\ \text{m s}^{-2}$).
Numerical value type. Enter your answer.
Answer: 10
For a solid cylinder $\dfrac{k^2}{R^2} = \dfrac12$:
$$a = \frac{g\sin\theta}{1 + \frac{k^2}{R^2}} = \frac23\times10\times\frac{\sqrt3}{2} = \frac{10}{\sqrt3}\ \text{m s}^{-2}$$
So $x = 10$.
Solution by Sreeraj P, M.Sc Physics