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Physics · Horizontal Circular Motion

JEE Main 2025 — 24 January, Morning Shift — Question 56

A car of mass ' mm ' moves on a banked road having radius ' r ' and banking angle θ\theta. To avoid slipping from banked road, the maximum permissible speed of the car is v0\mathrm{v}_{0}. The coefficient of friction μ\mu between the wheels of the car and the banked road is :-

  1. Option A:

    μ=v02+rgtan⁡θrg−v02tan⁡θ\mu=\frac{v_{0}^{2}+r g \tan \theta}{r g-v_{0}^{2} \tan \theta}

  2. Option B:

    μ=v02+rgtan⁡θrg+v02tan⁡θ\mu=\frac{\mathrm{v}_{0}^{2}+\mathrm{rg} \tan \theta}{\mathrm{rg}+\mathrm{v}_{0}^{2} \tan \theta}

  3. Option C:

    μ=v02−rgtan⁡θrg+v02tan⁡θ\mu=\frac{v_{0}^{2}-r g \tan \theta}{r g+v_{0}^{2} \tan \theta}

    Correct
  4. Option D:

    μ=v02−rgtan⁡θrg⁡−v02tan⁡θ\mu=\frac{v_{0}^{2}-r g \tan \theta}{\operatorname{rg}-v_{0}^{2} \tan \theta}

Answer: C

Step-by-step solution

Nsin⁡θ+fcos⁡θ=mv2R\mathrm{N} \sin \theta+\mathrm{f} \cos \theta=\frac{\mathrm{mv}^{2}}{\mathrm{R}}

Ncos⁡θ−fsin⁡θ=mg\mathrm{N} \cos \theta-\mathrm{f} \sin \theta=\mathrm{mg}

sin⁡θ+μcos⁡θcos⁡θ−μsin⁡θ=v2Rg\frac{\sin \theta+\mu \cos \theta}{\cos \theta-\mu \sin \theta}=\frac{\mathrm{v}^{2}}{R g}

Rgtan⁡θ+μRg⁡=v2−v2μtan⁡θ\operatorname{Rgtan} \theta+\mu \operatorname{Rg}=v^{2}-v^{2} \mu \tan \theta

μ=v2−Rgtan⁡θRg+v2tan⁡θ\mu=\frac{v^{2}-R g \tan \theta}{R g+v^{2} \tan \theta}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
24 January, Morning Shift
Subject
Physics
Chapter
Horizontal Circular Motion
Topic
Banking of Roads
A car of mass ' m ' moves on a banked road having radius ' r ' and… | JEE Main 2025 PYQ with Solution · DhiX AI