Q53 / 75 · JEE Main 2026 (04 April Shift 1) · Physics
The time taken by a block of mass $m$ to slide down from the highest point to the lowest point on a rough inclined plane is $50\%$ more compared to the time taken by the same block on identical inclined smooth plane. Both inclined planes are at $45^\circ$ with the horizontal. The coefficient of kinetic friction between the rough inclined surface and block is _____.
- A$3/4$
- B$2/3$
- Correct answerC$5/9$
- D$4/9$
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 $t_1 = \sqrt{\dfrac{2s}{a_1}} = \sqrt{\dfrac{2s}{g \sin \theta}}$.
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 $t_2 = \sqrt{\dfrac{2s}{a_2}} = \sqrt{\dfrac{2s}{g (\sin \theta - \mu_k \cos \theta)}}$.
Given that $t_2$ is $50\%$ more than $t_1$, we have $t_2 = t_1 + 0.5 t_1 = \dfrac{3}{2} t_1$.
Squaring both sides gives $t_2^2 = \dfrac{9}{4} t_1^2$.
Substituting the expressions for $t_1^2$ and $t_2^2$:
$\dfrac{2s}{g (\sin \theta - \mu_k \cos \theta)} = \dfrac{9}{4} \left( \dfrac{2s}{g \sin \theta} \right)$
$\dfrac{1}{\sin \theta - \mu_k \cos \theta} = \dfrac{9}{4 \sin \theta}$
$4 \sin \theta = 9 \sin \theta - 9 \mu_k \cos \theta$
$9 \mu_k \cos \theta = 5 \sin \theta$
$\mu_k = \dfrac{5}{9} \tan \theta$
Given $\theta = 45^\circ$, we have $\tan 45^\circ = 1$.
$\mu_k = \dfrac{5}{9} \times 1 = \dfrac{5}{9}$
Answer: $5/9$
For the smooth inclined plane, the acceleration of the block is $a_1 = g \sin \theta$.
The time taken to slide down is $t_1 = \sqrt{\dfrac{2s}{a_1}} = \sqrt{\dfrac{2s}{g \sin \theta}}$.
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 $t_2 = \sqrt{\dfrac{2s}{a_2}} = \sqrt{\dfrac{2s}{g (\sin \theta - \mu_k \cos \theta)}}$.
Given that $t_2$ is $50\%$ more than $t_1$, we have $t_2 = t_1 + 0.5 t_1 = \dfrac{3}{2} t_1$.
Squaring both sides gives $t_2^2 = \dfrac{9}{4} t_1^2$.
Substituting the expressions for $t_1^2$ and $t_2^2$:
$\dfrac{2s}{g (\sin \theta - \mu_k \cos \theta)} = \dfrac{9}{4} \left( \dfrac{2s}{g \sin \theta} \right)$
$\dfrac{1}{\sin \theta - \mu_k \cos \theta} = \dfrac{9}{4 \sin \theta}$
$4 \sin \theta = 9 \sin \theta - 9 \mu_k \cos \theta$
$9 \mu_k \cos \theta = 5 \sin \theta$
$\mu_k = \dfrac{5}{9} \tan \theta$
Given $\theta = 45^\circ$, we have $\tan 45^\circ = 1$.
$\mu_k = \dfrac{5}{9} \times 1 = \dfrac{5}{9}$
Answer: $5/9$
Hence, the correct answer is option: 'C'.