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

<p>$N \cos\theta - \mu N \sin\theta = mg$</p> <p>$N \sin\theta + \mu N \cos\theta = \frac{mv_0^2}{r}$</p> <p>Dividing the second equation by the first gives</p> <p>$$ \frac{v_0^2}{r} = \frac{g\left(\sin\theta + \mu \cos\theta\right)}{\cos\theta - \mu \sin\theta}. $$</p> <p>Multiplying both sides by $\cos\theta - \mu\sin\theta$ leads to</p> <p>$\frac{v_0^2}{r}(\cos\theta - \mu \sin\theta) = g(\sin\theta + \mu \cos\theta).$</p> <p>Expanding and grouping the terms with $\mu$:</p> <p>$$ \frac{v_0^2}{r}\cos\theta - g\sin\theta = \mu\left(\frac{v_0^2}{r}\sin\theta + g\cos\theta\right). $$</p> <p>Thus, solving for $\mu$:</p> <p>$$ \mu = \frac{\frac{v_0^2}{r}\cos\theta - g\sin\theta}{\frac{v_0^2}{r}\sin\theta + g\cos\theta}. $$</p> <p>Multiplying the numerator and the denominator by $r$ gives</p> <p>$\mu = \frac{v_0^2\cos\theta - rg\sin\theta}{v_0^2\sin\theta + rg\cos\theta}.$</p> <p>Dividing both the numerator and denominator by $\cos\theta$ (assuming $\cos\theta \neq 0$):</p> <p>$\mu = \frac{v_0^2 - rg\tan\theta}{v_0^2\tan\theta + rg}.$</p> <p>This result matches the expression</p> <p>$\mu=\frac{v_0^2 - rg\,\tan\theta}{rg + v_0^2\,\tan\theta}.$</p> <p>Thus, the correct answer is:</p> <p><strong>Option C</strong></p>