JEE Main 2026 — Laws of Motion Question with Solution

Let the length of the inclined plane be $s$.

For the smooth inclined plane, the acceleration of the block is $a_1=g\sin \theta$.

The time taken to slide down is $t_1=\sqrt{\frac{2s}{a_1}}=\sqrt{\frac{2s}{g\sin \theta }}$.

For the rough inclined plane, the acceleration of the block is $a_2=g\sin \theta -{\mu }_{k}g\cos \theta$.

The time taken to slide down is $t_2=\sqrt{\frac{2s}{a_2}}=\sqrt{\frac{2s}{g(\sin \theta -{\mu }_k\cos \theta )}}$.

Given that $t_2$ is $50\%$ more than $t_1$, we have $t_2=t_1+0.5t_1=\frac{3}{2}{t}_1$.

Squaring both sides gives $t_2^2=\frac{9}{4}{t}_1^2$.

Substituting the expressions for $t_1^2$ and $t_2^2$:

$\frac{2s}{g(\sin \theta -{\mu }_k\cos \theta )}=\frac{9}{4}\left(\frac{2s}{g\sin \theta }\right)$

$\frac{1}{\sin \theta -{\mu }_k\cos \theta }=\frac{9}{4\sin \theta }$

$4\sin \theta =9\sin \theta -9{\mu }_k\cos \theta$

$9{\mu }_k\cos \theta =5\sin \theta$

${\mu }_k=\frac{5}{9}\tan \theta$

Given $\theta =45^{\circ }$, we have $\tan 45^{\circ }=1$.

${\mu }_k=\frac{5}{9}\times 1=\frac{5}{9}$

Answer: $5/9$