JEE Physics 2025 Question with Solution

Given: A car of mass $m$ moves on a banked road of radius $r$ with banking angle $\theta$. The maximum permissible speed is $v_0$.

Find: The coefficient of friction $\mu$.

At the maximum permissible speed, the car is at the verge of slipping, so friction is at its limiting value. The required centripetal force is

$$F_c = \frac{mv_0^2}{r}$$

The forces acting are weight $mg$ downward, normal reaction $N$ perpendicular to the road, and friction acting along the slope.

Using the force balance written in the solution:

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

From the first equation,

$$N = \frac{mg}{\cos \theta + \mu \sin \theta}$$

Substitute this into the horizontal equation:

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

On simplifying, we obtain

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

Therefore, the correct option is C.

Given: A car moves on a banked road with radius $r$, banking angle $\theta$, and maximum permissible speed $v_0$.

Find: The coefficient of friction $\mu$.

The first approach resolves forces parallel and perpendicular to the incline.

Perpendicular balance:

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

So,

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

Parallel balance:

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

Substituting the value of $N$:

$$\left( \frac{mg + \frac{mv_0^2}{r} \sin \theta}{\cos \theta} \right) \sin \theta - \frac{mv_0^2}{r} \cos \theta = \mu \left( \frac{mg + \frac{mv_0^2}{r} \sin \theta}{\cos \theta} \right) \cos \theta$$

Simplifying gives

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

Hence, the coefficient of friction is $\frac{v_0^2 - rg \tan \theta}{rg + v_0^2 \tan \theta}$, so the correct option is C.