A block takes t time to slide… - JEE Main 2026 Physics PYQ

JEE Main 2026 — Physics Mechanics

2026integermedium

A block takes $t$ time to slide down a plane inclined at $45°$ to the horizontal. If the surface is made smooth (frictionless), the block takes time $\dfrac{t}{2}$ to slide down the plane. The coefficient of friction between the block and the inclined plane is $\left(\dfrac{\alpha}{100}\right)$ . The value of $\alpha$ is __________.

Official previous-year question

Held on 6 Apr 2026 · Verified 6 Jul 2026.

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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 is $t_1 = \dfrac{t}{2}$ . Using the second equation of motion $s = ut + \dfrac{1}{2}at^2$ with $u = 0$ :

$s = \dfrac{1}{2} (g \sin \theta) \left(\dfrac{t}{2}\right)^2 = \dfrac{1}{8} g \sin \theta t^2$

For the rough inclined plane, the acceleration of the block is $a_2 = g(\sin \theta - \mu \cos \theta)$ .

The time taken is $t_2 = t$ . Using the equation of motion again:

$s = \dfrac{1}{2} g(\sin \theta - \mu \cos \theta) t^2$

Equating the two expressions for $s$ :

$\dfrac{1}{8} g \sin \theta t^2 = \dfrac{1}{2} g(\sin \theta - \mu \cos \theta) t^2$

$\dfrac{1}{4} \sin \theta = \sin \theta - \mu \cos \theta$

Given $\theta = 45^{\circ}$ , we have $\sin 45^{\circ} = \cos 45^{\circ} = \dfrac{1}{\sqrt{2}}$ . Substituting these values:

$\dfrac{1}{4} = 1 - \mu$

$\mu = 1 - \dfrac{1}{4} = \dfrac{3}{4} = 0.75$

We are given $\mu = \dfrac{\alpha}{100}$ .

$\dfrac{\alpha}{100} = 0.75 \Rightarrow \alpha = 75$

Answer: $75$

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JEE Main 2026 — Physics Mechanics

2026integermedium

Official previous-year question

Held on 6 Apr 2026 · Verified 6 Jul 2026.

Did you get this right?

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 is $t_1 = \dfrac{t}{2}$ . Using the second equation of motion $s = ut + \dfrac{1}{2}at^2$ with $u = 0$ :

$s = \dfrac{1}{2} (g \sin \theta) \left(\dfrac{t}{2}\right)^2 = \dfrac{1}{8} g \sin \theta t^2$

For the rough inclined plane, the acceleration of the block is $a_2 = g(\sin \theta - \mu \cos \theta)$ .

The time taken is $t_2 = t$ . Using the equation of motion again:

$s = \dfrac{1}{2} g(\sin \theta - \mu \cos \theta) t^2$

Equating the two expressions for $s$ :

$\dfrac{1}{8} g \sin \theta t^2 = \dfrac{1}{2} g(\sin \theta - \mu \cos \theta) t^2$

$\dfrac{1}{4} \sin \theta = \sin \theta - \mu \cos \theta$

Given $\theta = 45^{\circ}$ , we have $\sin 45^{\circ} = \cos 45^{\circ} = \dfrac{1}{\sqrt{2}}$ . Substituting these values:

$\dfrac{1}{4} = 1 - \mu$

$\mu = 1 - \dfrac{1}{4} = \dfrac{3}{4} = 0.75$

We are given $\mu = \dfrac{\alpha}{100}$ .

$\dfrac{\alpha}{100} = 0.75 \Rightarrow \alpha = 75$

Answer: $75$

Your note

JEE Main Mechanics in other years

Mechanics 2025 Mechanics 2024 Mechanics 2023 Mechanics 2022 Mechanics 2021 Mechanics 2020 Mechanics 2019 Mechanics 2018 Mechanics 2017 Mechanics 2016 Mechanics 2015 Mechanics 2014

Keep practicing Mechanics

Work through every JEE Main Mechanics PYQ, year by year.

Browse all Physics topics

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Solution

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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