Q 11-06-155JEE MainJEE Main 2021 (20 Jul, Shift 1)Easy
A circular disc reaches from top to bottom of an inclined plane of length $L$. When it slips down the plane, it takes time $t_1$. When it rolls down the plane, it takes time $t_2$. The value of $\dfrac{t_2}{t_1}$ is $\sqrt{\dfrac3x}$. The value of $x$ will be ______.
Numerical value type. Enter your answer.
Answer: 2
Slipping (no friction): $a_1 = g\sin\theta$. Rolling disc: $a_2 = \dfrac{g\sin\theta}{1 + \frac12} = \dfrac23g\sin\theta$.
$t = \sqrt{\dfrac{2L}{a}}$, so $\dfrac{t_2}{t_1} = \sqrt{\dfrac{a_1}{a_2}} = \sqrt{\dfrac32}$, giving $x = 2$.
Solution by Sreeraj P, M.Sc Physics