JEE Main 2025 — Circular Motion Question with Solution

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

$$N\sin \theta +\mu N\cos \theta =\frac{m{v}_0^2}{r}$$

Dividing the second equation by the first gives

$$\frac{v_0^2}{r}=\frac{g\left(\sin \theta +\mu \cos \theta \right)}{\cos \theta -\mu \sin \theta }.$$

Multiplying both sides by $\cos \theta -\mu \sin \theta$ leads to

$$\frac{v_0^2}{r}(\cos \theta -\mu \sin \theta )=g(\sin \theta +\mu \cos \theta ).$$

Expanding and grouping the terms with $\mu$:

$$\frac{v_0^2}{r}\cos \theta -g\sin \theta =\mu \left(\frac{v_0^2}{r}\sin \theta +g\cos \theta \right).$$

Thus, solving for $\mu$:

$$\mu =\frac{\frac{v_0^2}{r}\cos \theta -g\sin \theta }{\frac{v_0^2}{r}\sin \theta +g\cos \theta }.$$

Multiplying the numerator and the denominator by $r$ gives

$$\mu =\frac{v_0^2\cos \theta -rg\sin \theta }{v_0^2\sin \theta +rg\cos \theta }.$$

Dividing both the numerator and denominator by $\cos \theta$ (assuming $\cos \theta \ne 0$):

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

This result matches the expression

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

Thus, the correct answer is:

Option C