A given object takes n times the time to slide down 45 ° rough inclined plane as it takes the time to slide down an identical perfectly smooth 45 ° inclined plane. The coefficient of kinetic friction between the object and the surface of inclined plane is :

Physics · Friction

JEE Main 2024 — 8 April, Shift 2 — Question 41

A given object takes $n$ times the time to slide down $45^{\circ}$ rough inclined plane as it takes the time to slide down an identical perfectly smooth $45^{\circ}$ inclined plane. The coefficient of kinetic friction between the object and the surface of inclined plane is :

Option A:
$1-\frac{1}{\mathrm{n}^{2}}$
Correct
Option B:
$1-\mathrm{n}^{2}$
Option C:
$\sqrt{1-\frac{1}{\mathrm{n}^{2}}}$
Option D:
$\sqrt{1-\mathrm{n}^{2}}$

Answer: A

Step-by-step solution

Case-1 : No friction

\begin{aligned} & \mathrm{a}=\mathrm{g} \sin \theta & \ell=\frac{1}{2}(\mathrm{~g} \sin \theta) \mathrm{t}_{1}^{2} & \mathrm{t}_{1}=\sqrt{\frac{2 \ell}{\mathrm{~g} \sin \theta}} \end{aligned}

Case-2 : With friction

$\mathrm{a}=\mathrm{g} \sin \theta-\mu \mathrm{g} \cos \theta$

$\ell=\frac{1}{2}(\mathrm{~g} \sin \theta-\mu \mathrm{g} \cos \theta) \mathrm{t}_{2}^{2}$

$\sqrt{\frac{2 \ell}{\mathrm{~g} \sin \theta-\mu \mathrm{g} \cos \theta}}=\mathrm{n} \sqrt{\frac{2 \ell}{\mathrm{~g} \sin \theta}}$

$\mu=1-\frac{1}{\mathrm{n}^{2}}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 8 April, Shift 2
Subject: Physics
Chapter: Friction
Topic: Single Block Problems Involving Friction