JEE Mains · Physics · STD 11 - Friction
- JEE Mains 2025
- Medium
- MCQ
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
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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, \(N=m g \cos \theta+\frac{m v_0^2}{r} \sin \theta\)
\(f_r=\mu m g \cos \theta+\frac{\mu m v_0^2}{r} \sin \theta\)
\(\begin{aligned}
& \text { And } \frac{m v_0^2}{r} \cos \theta=m g \sin \theta+f_r \\
& \Rightarrow \frac{m v_0^2}{r} \cos \theta-m g \sin \theta=\mu\left(m g \cos \theta+\frac{m v_0^2}{r} \sin \theta\right) \\
& \Rightarrow\left(v_0^2-g r \tan \theta\right)=\mu\left(v_0^2 \tan \theta+g r\right) \\
& \Rightarrow \mu=\frac{v_0^2-g r \tan \theta}{g r+v_0^2 \tan \theta}
\end{aligned}\)
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