Torque, Couple & Equilibrium of Rigid Bodies question from NEET 2025

Answer: (d)

1. The wall is smooth, so it gives only a horizontal push $N_w$. Horizontal balance: friction $f = N_w$.
2. Rod makes $60^\circ$ with the wall: top height $= L\cos 60^\circ$, CM horizontal distance from foot $= \frac{L}{2}\sin 60^\circ$.
3. Moments about the foot: $N_w L\cos 60^\circ = mg\,\frac{L}{2}\sin 60^\circ \Rightarrow N_w = \frac{mg}{2}\tan 60^\circ$
4. $f = \frac{200}{2} \times \sqrt{3} = 100\sqrt{3}$ N
5. So the answer is (d) $100\sqrt{3}$ N.