JEE Mains · Physics · STD 11 - 4.2 friction
A car of mass ' \(m\) ' moves on a banked road having radius ' \(r\) ' and banking angle \(\theta\). To avoid slipping from banked road, the maximum permissible speed of the car is \(v_0\). The coefficient of friction \(\mu\) between the wheels of the car and the banked road is
- A \(\mu=\frac{\mathrm{v}_{\mathrm{o}}^2+\mathrm{rg} \tan \theta}{\mathrm{rg}+\mathrm{v}_{\mathrm{o}}^2 \tan \theta}\)
- B \(\mu=\frac{v_0^2-r g \tan \theta}{r g-v_0^2 \tan \theta}\)
- C \(\mu=\frac{v_0^2-r g \tan \theta}{r g+v_0^2 \tan \theta}\)
- D \(\mu=\frac{v_o^2+r g \tan \theta}{\mathrm{rg}-\mathrm{v}_{\mathrm{o}}^2 \tan \theta}\)
Answer & Solution
Correct Answer
(C) \(\mu=\frac{v_0^2-r g \tan \theta}{r g+v_0^2 \tan \theta}\)
Step-by-step Solution
Detailed explanation
So, \(N=m g \cos \theta+\frac{m v_0^2}{r} \sin \theta\) \(f_r=\mu m g \cos \theta+\frac{\mu m v_0^2}{r} \sin \theta\)…
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