A car is negotiating a curved road of radius \mathbf{r}. The road is banked at an 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 \quad \sqrt{\operatorname{gR}^{2} \frac{\mu_{\mathrm{s}}+\tan \theta}{1-\mu_{\mathrm{s}} \tan \theta}}
B \sqrt{\operatorname{gR} \frac{\mu_{\mathrm{s}}+\tan \theta}{1-\mu_{\mathrm{s}} \tan \theta}}
C \sqrt{\frac{\mathrm{g}}{\mathrm{R}} \frac{\mu_{\mathrm{s}}+\tan \theta}{1-\mu_{\mathrm{s}} \tan \theta}}
D \sqrt{\frac{\mathrm{g}}{\mathrm{R}^{2}} \frac{\mu_{\mathrm{s}}+\tan \theta}{1-\mu_{\mathrm{s}} \tan \theta}}
A car is negotiating a curved road of radius \mathbf{r}. The road is banked at an 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 \quad \sqrt{\operatorname{gR}^{2} \frac{\mu_{\mathrm{s}}+\tan \theta}{1-\mu_{\mathrm{s}} \tan \theta}}
B \sqrt{\operatorname{gR} \frac{\mu_{\mathrm{s}}+\tan \theta}{1-\mu_{\mathrm{s}} \tan \theta}}
C \sqrt{\frac{\mathrm{g}}{\mathrm{R}} \frac{\mu_{\mathrm{s}}+\tan \theta}{1-\mu_{\mathrm{s}} \tan \theta}}
D \sqrt{\frac{\mathrm{g}}{\mathrm{R}^{2}} \frac{\mu_{\mathrm{s}}+\tan \theta}{1-\mu_{\mathrm{s}} \tan \theta}}

Correct Option B

Solution:

For vertical equilibrium on the curved road we have,

N \cos \theta=m g+f_{1} \operatorname{Sin} \theta----(1)

Forhorizontal equilibrium on the curved road we have,

\mathrm{NSin} \theta+f_{1} \operatorname{Cos} \theta=\frac{m v^{2}}{r}---(2)

Now solving these two equatiopns we get

\sqrt{g R \frac{\mu_{s}+\tan \theta}{1-\mu_{s} \tan \theta}}