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A car is negotiating a curved road of radius R. The road is banked at angle $\theta$. The coefficient of friction between the tyres of the car and the road is $\mu_s$. The maximum safe velocity on this road is
A
$\sqrt{gR\left(\frac{\mu_s + \tan\theta}{1 - \mu_s\tan\theta}\right)}$
B
$\sqrt{\frac{g}{R}\left(\frac{\mu_s + \tan\theta}{1 - \mu_s\tan\theta}\right)}$
C
$\sqrt{\frac{g}{R^2}\left(\frac{\mu_s + \tan\theta}{1 - \mu_s\tan\theta}\right)}$
D
$\sqrt{gR^2\left(\frac{\mu_s + \tan\theta}{1 - \mu_s\tan\theta}\right)}$
Explanation
Friction acts down the slope at maximum speed.
Detailed Solution
Vertical: $N\cos\theta = mg + f\sin\theta \Rightarrow mg = N\cos\theta - f\sin\theta$ ...(i)
Horizontal: $N\sin\theta + f\cos\theta = \frac{mv^2}{R}$ ...(ii)
Dividing (ii) by (i) with $f = \mu_sN$: $\frac{v^2}{Rg} = \frac{\sin\theta + \mu_s\cos\theta}{\cos\theta - \mu_s\sin\theta}$
$v = \sqrt{Rg\left[\frac{\tan\theta + \mu_s}{1 - \mu_s\tan\theta}\right]}$