A car of mass ' $m$ ' moves on a banked road having radius ' $r$ ' and banking angle $\theta$. To avoid…

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
  1. $\mu=\frac{\mathrm{v}_{\mathrm{o}}^2+\mathrm{rg} \tan \theta}{\mathrm{rg}+\mathrm{v}_{\mathrm{o}}^2 \tan \theta}$
  2. $\mu=\frac{v_0^2-r g \tan \theta}{r g-v_0^2 \tan \theta}$
  3. $\mu=\frac{v_0^2-r g \tan \theta}{r g+v_0^2 \tan \theta}$
  4. $\mu=\frac{v_o^2+r g \tan \theta}{\mathrm{rg}-\mathrm{v}_{\mathrm{o}}^2 \tan \theta}$

Solution


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

Asked in: JEE Main 2025 (24 Jan Shift 1)

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