JEE Main 2026 — Physics Mechanics

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

For the smooth inclined plane, the acceleration of the block is $a_1 = g \sin \theta$a1​=gsinθ.

The time taken to slide down is $t_1 = \sqrt{\dfrac{2s}{a_1}} = \sqrt{\dfrac{2s}{g \sin \theta}}$t1​=a1​2s​​=gsinθ2s​​.

For the rough inclined plane, the acceleration of the block is $a_2 = g \sin \theta - \mu_k g \cos \theta$a2​=gsinθ−μk​gcosθ.

The time taken to slide down is $t_2 = \sqrt{\dfrac{2s}{a_2}} = \sqrt{\dfrac{2s}{g (\sin \theta - \mu_k \cos \theta)}}$t2​=a2​2s​​=g(sinθ−μk​cosθ)2s​​.

Given that $t_2$t2​ is $50\%$50% more than $t_1$t1​, we have $t_2 = t_1 + 0.5 t_1 = \dfrac{3}{2} t_1$t2​=t1​+0.5t1​=23​t1​.

Squaring both sides gives $t_2^2 = \dfrac{9}{4} t_1^2$t22​=49​t12​.

Substituting the expressions for $t_1^2$t12​ and $t_2^2$t22​:

$$\dfrac{2s}{g (\sin \theta - \mu_k \cos \theta)} = \dfrac{9}{4} \left( \dfrac{2s}{g \sin \theta} \right)$$g(sinθ−μk​cosθ)2s​=49​(gsinθ2s​)

$$\dfrac{1}{\sin \theta - \mu_k \cos \theta} = \dfrac{9}{4 \sin \theta}$$sinθ−μk​cosθ1​=4sinθ9​

$$4 \sin \theta = 9 \sin \theta - 9 \mu_k \cos \theta$$4sinθ=9sinθ−9μk​cosθ

$$9 \mu_k \cos \theta = 5 \sin \theta$$9μk​cosθ=5sinθ

$$\mu_k = \dfrac{5}{9} \tan \theta$$μk​=95​tanθ

Given $\theta = 45^\circ$θ=45∘, we have $\tan 45^\circ = 1$tan45∘=1.

$$\mu_k = \dfrac{5}{9} \times 1 = \dfrac{5}{9}$$μk​=95​×1=95​

Answer: $5/9$5/9