Q 11-04-166JEE MainJEE Main 2021 (20 Jul, Shift 2)Medium
A body of mass $m$ is launched up on a rough inclined plane making an angle of $30^\circ$ with the horizontal. The coefficient of friction between the body and plane is $\dfrac{\sqrt x}{5}$ if the time of ascent is half of the time of descent. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 3
Up the plane: $a_1 = g(\sin30^\circ + \mu\cos30^\circ)$. Down: $a_2 = g(\sin30^\circ - \mu\cos30^\circ)$.
Same distance: $a_1t_1^2 = a_2t_2^2$ with $t_2 = 2t_1$, so $a_1 = 4a_2$:
$$\frac12 + \frac{\sqrt3}{2}\mu = 4\left(\frac12 - \frac{\sqrt3}{2}\mu\right) \Rightarrow \frac{5\sqrt3}{2}\mu = \frac32 \Rightarrow \mu = \frac{\sqrt3}{5}$$
So $x = 3$.
Solution by Sreeraj P, M.Sc Physics