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JEE Main 202524 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
24 January
Shift
Morning Shift
Subject
Physics
Chapter
Laws of Motion

JEE Main 2025 · 24 January · Morning Shift · Physics · Laws of Motion

A car of mass ' m ' moves on a banked road having radius ' r ' and...

A car of mass ' mm ' moves on a banked road having radius ' rr ' and banking angle θ\theta. To avoid slipping from banked road, the maximum permissible speed of the car is v0v_0. The coefficient of friction μ\mu between the wheels of the car and the banked road is

Options

  1. Aμ=vo2+rgtan⁡θrg+vo2tan⁡θ\mu=\frac{\mathrm{v}_{\mathrm{o}}^2+\mathrm{rg} \tan \theta}{\mathrm{rg}+\mathrm{v}_{\mathrm{o}}^2 \tan \theta}
  2. Bμ=v02−rgtan⁡θrg−v02tan⁡θ\mu=\frac{v_0^2-r g \tan \theta}{r g-v_0^2 \tan \theta}
  3. Cμ=v02−rgtan⁡θrg+v02tan⁡θ\mu=\frac{v_0^2-r g \tan \theta}{r g+v_0^2 \tan \theta}
  4. Dμ=vo2+rgtan⁡θrg−vo2tan⁡θ\mu=\frac{v_o^2+r g \tan \theta}{\mathrm{rg}-\mathrm{v}_{\mathrm{o}}^2 \tan \theta}

Correct answer

C. μ=v02−rgtan⁡θrg+v02tan⁡θ\mu=\frac{v_0^2-r g \tan \theta}{r g+v_0^2 \tan \theta}

Step-by-step solution

So, N=mgcos⁡θ+mv02rsin⁡θfr=μmgcos⁡θ+μmv02rsin⁡θN=m g \cos \theta+\frac{m v_0^2}{r} \sin \theta f_r=\mu m g \cos \theta+\frac{\mu m v_0^2}{r} \sin \theta And  mv02rcos⁡θ=mgsin⁡θ+fr⇒mv02rcos⁡θ−mgsin⁡θ=μ(mgcos⁡θ+mv02rsin⁡θ)⇒(v02−grtan⁡θ)=μ(v02tan⁡θ+gr)⇒μ=v02−grtan⁡θgr+v02tan⁡θ\begin{aligned} & \text{\text{ And } } \frac{m v_0^2}{r} \cos \theta=m g \sin \theta+f_r \\ & \Rightarrow \frac{m v_0^2}{r} \cos \theta-m g \sin \theta=\mu\left(m g \cos \theta+\frac{m v_0^2}{r} \sin \theta\right) \\ & \Rightarrow\left(v_0^2-g r \tan \theta\right)=\mu\left(v_0^2 \tan \theta+g r\right) \\ & \Rightarrow \mu=\frac{v_0^2-g r \tan \theta}{g r+v_0^2 \tan \theta} \end{aligned}$$
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