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

  $\mu=\frac{v_0^2-r g \tan \theta}{r g-v_0^2 \tan \theta}$
• Correct answer
  C

  $\mu=\frac{v_0^2-r g \tan \theta}{r g+v_0^2 \tan \theta}$
• D

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

Solution

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

Hence, the correct answer is option: 'C'.