JEE Main 2025 Apr 7 Shift 1, Physics Q14: Friction

Newton's second law along the incline (taking down-the-slope positive):

$mg\sin 60^{\circ }-\mu mg\cos 60^{\circ }=m\cdot \frac{g}{2}$

Cancel $m$ and divide by $g$:

$\sin 60^{\circ }-\mu \cos 60^{\circ }=\frac{1}{2}$

Substitute $\sin 60^{\circ }=\frac{\sqrt{3}}{2}$ and $\cos 60^{\circ }=\frac{1}{2}$:

$\frac{\sqrt{3}}{2}-\frac{\mu }{2}=\frac{1}{2}$

$\Rightarrow \mu =\sqrt{3}-1$

$\therefore$ Option (1).