The time taken by a block of… - JEE Main 2026 Physics PYQ

JEE Main 2026 — Physics Mechanics

2026mcqmedium

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 _____.

Official previous-year question

Held on 4 Apr 2026 · Verified 6 Jul 2026.

Options

A
$3/4$
B
$2/3$
C
$5/9$
D
$4/9$

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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: $5/9$

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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: $5/9$

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Options

Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics