JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Laws of Motion

On a banked road inclined at angle $\theta$, let the maximum safe speed be $v_0$. At this speed the car is on the verge of sliding outward (up the incline), so friction acts down the incline at its limiting value $f=\mu N$.

Vertical equilibrium (net vertical force is zero):
$N\cos \theta =mg+f\sin \theta$
$N(\cos \theta -\mu \sin \theta )=mg$ ... (1)

Horizontal direction supplies the centripetal force:
$N\sin \theta +f\cos \theta =\frac{m{v}_0^2}{r}$
$N(\sin \theta +\mu \cos \theta )=\frac{m{v}_0^2}{r}$ ... (2)

Dividing (2) by (1):
$\frac{\sin \theta +\mu \cos \theta }{\cos \theta -\mu \sin \theta }=\frac{v_0^2}{rg}$

Dividing numerator and denominator on the left by $\cos \theta$:
$\frac{\tan \theta +\mu }{1-\mu \tan \theta }=\frac{v_0^2}{rg}$

Cross-multiplying:
$rg(\tan \theta +\mu )=v_0^2(1-\mu \tan \theta )$
$rg\tan \theta +\mu rg=v_0^2-\mu v_0^2\tan \theta$
$\mu (rg+v_0^2\tan \theta )=v_0^2-rg\tan \theta$

$\mu =\frac{v_0^2-rg\tan \theta }{rg+v_0^2\tan \theta }$