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JEE Main 2019 · Physics · Kinematics and Laws of Motion · medium

A car travels on a rough circular road of radius 45 m that is banked at an angle θ with tan θ=0.75. The coefficient of friction between the tyres and…

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A car travels on a rough circular road of radius 45 m45\ \mathrm{m} that is banked at an angle θ\theta with tan⁡θ=0.75\tan\theta=0.75. The coefficient of friction between the tyres and the road is 0.50.5. Take g=10 m s−2g=10\ \mathrm{m\,s^{-2}}. The maximum speed with which the car can go round the curve without skidding outward is
  1. A. 15 m s−115\ \mathrm{m\,s^{-1}}
  2. B. 18.4 m s−118.4\ \mathrm{m\,s^{-1}}
  3. C. 30 m s−130\ \mathrm{m\,s^{-1}}
  4. D. 36 m s−136\ \mathrm{m\,s^{-1}}
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Answer: (C) 30 m s−130\ \mathrm{m\,s^{-1}}

At the limit, friction acts down the slope. Resolving along the horizontal (centripetal) and vertical directions: Nsin⁡θ+fcos⁡θ=mv2rN\sin\theta+f\cos\theta=\dfrac{mv^2}{r} and Ncos⁡θ−fsin⁡θ=mgN\cos\theta-f\sin\theta=mg with f=μNf=\mu N. Dividing, v2rg=tan⁡θ+μ1−μtan⁡θ=0.75+0.51−0.375=2\dfrac{v^2}{rg}=\dfrac{\tan\theta+\mu}{1-\mu\tan\theta}=\dfrac{0.75+0.5}{1-0.375}=2. So v2=2(45)(10)=900v^2=2(45)(10)=900 and v=30 m s−1v=30\ \mathrm{m\,s^{-1}}. Ignoring friction gives rgtan⁡θ≈18.4 m s−1\sqrt{rg\tan\theta}\approx18.4\ \mathrm{m\,s^{-1}}; a flat road would give μrg=15 m s−1\sqrt{\mu rg}=15\ \mathrm{m\,s^{-1}}.
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