Laws of Motion Part 2

Concept:For a flexible chain in equilibrium, each half is acted upon by its weight, tension at the support, and tension at the lowest point.Explanation:Consider the left half of the chain.Its mass is $\frac{m}{2}$, so its weight is $\frac{mg}{2}$ acting vertically downward.Let $T_2$ be the tension at the support, making an angle $30^\circ$ with the horizontal.Let $T_1$ be the tension at the lowest point, acting purely horizontally.For vertical equilibrium of the half‑chain:$T_2 \sin 30^\circ = \frac{mg}{2}$$\Rightarrow T_2 \times \frac{1}{2} = \frac{mg}{2}$$\Rightarrow T_2 = mg$For horizontal equilibrium:$T_1 = T_2 \cos 30^\circ$$\Rightarrow T_1 = mg \times \frac{\sqrt{3}}{2}$$\Rightarrow T_1 = \frac{\sqrt{3}}{2} mg$Answer:The tension at the lowest point is $\frac{\sqrt{3}}{2} mg$, which corresponds to option B.