A car is negotiating a curved road of radius $\text{R}$. The road is banked at an angle $\mathrm{\theta }$. The coefficient of friction between the tyres of the car and the road is ${\mathrm{\mu }}_{\mathrm{s}}$. The maximum safe velocity on this road is:

• A

  $\sqrt{g{\mathrm{R}}^2\frac{{\mathrm{\mu }}_{\mathrm{s}}+\tan \mathrm{\theta }}{1-{\mathrm{\mu }}_{\mathrm{s}}\tan \mathrm{\theta }}}$
• Correct answer
  B

  $\sqrt{g\mathrm{R}\frac{{\mathrm{\mu }}_{\mathrm{s}}+\tan \mathrm{\theta }}{1-{\mathrm{\mu }}_{\mathrm{s}}\tan \mathrm{\theta }}}$
• C

  $\sqrt{\frac{g}{\mathrm{R}}\frac{{\mathrm{\mu }}_{\mathrm{s}}+\tan \mathrm{\theta }}{1-{\mathrm{\mu }}_{\mathrm{s}}\tan \mathrm{\theta }}}$
• D

  $\sqrt{\frac{g}{{\mathrm{R}}^2}\frac{{\mathrm{\mu }}_{\mathrm{s}}+\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{\theta }}{1-{\mathrm{\mu }}_{\mathrm{s}}\tan \mathrm{\theta }}}$

Solution

$\mathrm{N}=\mathrm{m}\mathrm{g}\cos \mathrm{\theta }+\frac{\mathrm{m}{\mathrm{v}}^2}{\mathrm{R}}\sin \mathrm{\theta }$

$f_{\max }=\mu \mathrm{N}\Rightarrow$ $f_{\mathrm{m}\mathrm{a}\mathrm{x}}={\mu }_s\mathrm{m}\mathrm{g}\cos \mathrm{\theta }+\frac{{\mu }_s\mathrm{m}{\mathrm{v}}^2}{\mathrm{R}}\sin \mathrm{\theta }$

$\mathrm{mg}\sin \theta +f_{\max }=\frac{\mathrm{m}{\mathrm{v}}^2}{R}\cos \theta$

$\mathrm{m}\mathrm{g}\sin \mathrm{\theta }+{\mu }_s\mathrm{m}\mathrm{g}\cos \mathrm{\theta }+\frac{{\mu }_s\mathrm{m}{\mathrm{v}}^2}{\mathrm{R}}\sin \mathrm{\theta }=\frac{\mathrm{m}{\mathrm{v}}^2}{\mathrm{R}}\cos \mathrm{\theta }$

$\mathrm{g}\sin \mathrm{\theta }+{\mu }_s\mathrm{g}\cos \mathrm{\theta }=\frac{{\mathrm{v}}^2}{\mathrm{R}}\left(\cos \mathrm{\theta }-{\mu }_s\sin \mathrm{\theta }\right)$

$\mathrm{g}\mathrm{R}\left[\frac{\tan \mathrm{\theta }+{\mu }_s}{1-{\mu }_s\tan \mathrm{\theta }}\right]={\mathrm{v}}^2$

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