NEET (UG) 2025 — Question 21

The rod is in equilibrium under the forces: weight $mg = 200\,\text{N}$ acting at the centre, normal reaction from the wall $N_1$ (horizontal), normal reaction from the floor $N_2$ (vertical), and friction $f_s$ (horizontal) from the floor. Vertical equilibrium gives $N_2 = mg = 200\,\text{N}$. Take torque about point A (the point of contact with the floor). The torque due to $N_2$ and $f_s$ about A is zero because they pass through A. Torque due to $mg$: $mg \cdot \frac{L}{2} \sin 60^\circ = 200 \cdot \frac{5}{2} \cdot \frac{\sqrt{3}}{2} = 250\sqrt{3}\,\text{Nm}$ (clockwise). Torque due to $N_1$: $N_1 \cdot L \cos 60^\circ = N_1 \cdot 5 \cdot \frac{1}{2} = 2.5 N_1$ (counterclockwise). Setting clockwise torque equal to counterclockwise torque: $250\sqrt{3} = 2.5 N_1 \Rightarrow N_1 = 100\sqrt{3}\,\text{N}$. Horizontal equilibrium gives $f_s = N_1 = 100\sqrt{3}\,\text{N}$.