1. For maximum speed on a banked road

When the car moves with the maximum permissible speed $v_0$, it tends to slip up the bank. Therefore, friction acts down the slope.

2. Forces acting on the car

The forces are:

• Weight $mg$ vertically downward
• Normal reaction $N$ perpendicular to the road
• Friction $f = \mu N$ down the slope

Resolve forces into:

• Horizontal direction toward the center of the circle
• Vertical direction

3. Resolve forces horizontally

The horizontal components provide centripetal force:

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

Since $f=\mu N$,

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

$N(\sin\theta+\mu\cos\theta)=\frac{mv_0^2}{r} \quad ...(1)$

4. Resolve forces vertically

Vertical acceleration is zero, so:

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

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

$N(\cos\theta-\mu\sin\theta)=mg \quad ...(2)$

5. Divide (1) by (2)

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

Let

$x=\frac{v_0^2}{rg}$

Then

$\sin\theta+\mu\cos\theta=x(\cos\theta-\mu\sin\theta)$

Expand:

$\sin\theta+\mu\cos\theta=x\cos\theta-x\mu\sin\theta$

Bring $\mu$ terms together:

$\mu\cos\theta+x\mu\sin\theta=x\cos\theta-\sin\theta$

$\mu(\cos\theta+x\sin\theta)=x\cos\theta-\sin\theta$

$\mu=\frac{x\cos\theta-\sin\theta}{\cos\theta+x\sin\theta}$

Now divide numerator and denominator by $\cos\theta$:

$\mu=\frac{x-\tan\theta}{1+x\tan\theta}$

Substitute $x=\dfrac{v_0^2}{rg}$:

$\mu=\frac{\frac{v_0^2}{rg}-\tan\theta}{1+\frac{v_0^2}{rg}\tan\theta}$

Multiply numerator and denominator by $rg$:

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

6. Match with options

This matches:

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

So the correct option is C.