NEET (UG) 2025 — Question 4

Let $L$ be the length of each incline, $\theta = 45^\circ$. For the smooth surface, acceleration $a_s = g \sin\theta$. For the rough surface, acceleration $a_r = g(\sin\theta - \mu_k \cos\theta)$. Using $L = \frac{1}{2} a t^2$, we have $\frac{t_r}{t_s} = \sqrt{\frac{a_s}{a_r}}$. Given $t_r = 2 t_s$, so $\frac{t_r}{t_s} = 2$. Thus $2 = \sqrt{\frac{a_s}{a_r}}$ → $4 = \frac{a_s}{a_r}$. Substitute accelerations: $4 = \frac{g \sin\theta}{g(\sin\theta - \mu_k \cos\theta)}$. Simplify: $4 = \frac{\sin\theta}{\sin\theta - \mu_k \cos\theta}$. For $\theta = 45^\circ$, $\sin\theta = \cos\theta = \frac{1}{\sqrt{2}}$. Thus $4 = \frac{1/\sqrt{2}}{(1/\sqrt{2})(1 - \mu_k)}$ → $4 = \frac{1}{1 - \mu_k}$. Solving: $1 - \mu_k = \frac{1}{4}$ → $\mu_k = \frac{3}{4} = 0.75$. Hence the coefficient of kinetic friction is $0.75$, which corresponds to option D.