A vehicle is moving on a banked road, ‘ $\theta$ ’ is angle of banking and ‘ $\mu$ ’ is coefficient of friction between tyre of vehicle and road. The vehicle is moving along horizontal circular path of radius ($r$) velocity of vehicle for which it does not slip is ‘$v$’. Choose the correct alternative

Tendency of vehicle is to slip down

$N\cos \theta +f\sin \theta =mg$

$\mathrm{Nsin}\theta -\mathrm{fcos}\theta =\frac{{\mathrm{mV}}^2}{\mathrm{r}};\therefore \mathrm{f}=\mu \mathrm{N}$

$\mathrm{N}(\cos \theta +\mu \sin \theta )=\mathrm{mg}$

$\mathrm{N}(\sin \theta -\mu \cos \theta )=\frac{{\mathrm{mV}}^2}{\mathrm{r}}$

On solving we get $={\mathrm{V}}_{min}=\sqrt{\frac{\mathrm{rg}(\tan \theta -\mu )}{1+\mu \tan \theta }}$

Tendency of vehicle is to slip up

$N\cos \theta -f\sin \theta =mg$

$\mathrm{Nsin}\theta +f\cos \theta =\frac{m{v}^2}{r}$

$\therefore \mathrm{f}=\mu \mathrm{N}$

$\mathrm{N}(\cos \theta -\mu \sin \theta )=\mathrm{mg}$

$\mathrm{N}(\sin \theta +\mu \cos \theta )=\frac{{\mathrm{mV}}^2}{\mathrm{r}}$

on solving we get  ${\mathrm{V}}_{max}=\sqrt{\frac{\mathrm{rg}(\tan \theta +\mu )}{1-\mu \tan \theta }}$