QJEEM25Ph9e9f: JEE Main 2025 Physics - Class 11 JEEM25Ph9e9f Solution: A car of mass ' ' moves on a banked road having radius ' ' and…

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

$N \sin\theta + \mu N \cos\theta = \frac{mv_0^2}{r}$

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 $\cos\theta \neq 0$ ):

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

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