Question 197 of 210

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$\mu=\frac{v_o^2+r g \tan \theta}{\mathrm{rg}-\mathrm{v}_{\mathrm{o}}^2 \tan \theta}$

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Answer: (C) $\mu=\frac{v_0^2-r g \tan \theta}{r g+v_0^2 \tan \theta}$

So, $N=m g \cos \theta+\frac{m v_0^2}{r} \sin \theta$
$f_r=\mu m g \cos \theta+\frac{\mu m v_0^2}{r} \sin \theta$
$\begin{aligned}
& \text { And } \frac{m v_0^2}{r} \cos \theta=m g \sin \theta+f_r \\
& \Rightarrow \frac{m v_0^2}{r} \cos \theta-m g \sin \theta=\mu\left(m g \cos \theta+\frac{m v_0^2}{r} \sin \theta\right) \\
& \Rightarrow\left(v_0^2-g r \tan \theta\right)=\mu\left(v_0^2 \tan \theta+g r\right) \\
& \Rightarrow \mu=\frac{v_0^2-g r \tan \theta}{g r+v_0^2 \tan \theta}
\end{aligned}$