Physics · Friction

JEE Main 2026 — 6 April, Evening Shift — Question 16

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

Answer: 75

Numerical answer — enter this value.

Step-by-step solution

For rough: $a_1 = g\sin45 - \mu g\cos45 = g/\sqrt{2}(1-\mu)$ . For smooth: $a_2 = g/\sqrt{2}$ . Distance same: $\frac12 a_1 t^2 = \frac12 a_2 (t/2)^2 \Rightarrow a_1 = a_2/4$ . So $g/\sqrt{2}(1-\mu) = \frac{1}{4}\cdot g/\sqrt{2} \Rightarrow 1-\mu = 1/4 \Rightarrow \mu = 3/4 = 75/100$ , so $\alpha=75$ .

Answer key and solution verified before publishing.

Practise Friction

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2026
Paper: 6 April, Evening Shift
Subject: Physics
Chapter: Friction
Topic: Single Block Problems Involving Friction