JEE Main 2025 — Laws of Motion Question with Solution

So, $N=mg\cos \theta +\frac{m{v}_0^2}{r}\sin \theta$ $f_r=\mu mg\cos \theta +\frac{\mu m{v}_0^2}{r}\sin \theta$ $\begin{aligned} & \text { And } \frac{m v_0^2}{r} \cos \theta=m g \sin \theta+f_r \\ & \Rightarrow \frac{m v_0^2}{r} \cos \theta-m g \sin \theta=\mu\left(m g \cos \theta+\frac{m v_0^2}{r} \sin \theta\right) \\ & \Rightarrow\left(v_0^2-g r \tan \theta\right)=\mu\left(v_0^2 \tan \theta+g r\right) \\ & \Rightarrow \mu=\frac{v_0^2-g r \tan \theta}{g r+v_0^2 \tan \theta} \end{aligned}$