Question

A flexible chain of mass \(m\) hangs between two fixed points at the same level. The inclination of the chain with the horizontal at the two points of support is \(30^\circ\). Considering the equilibrium of each half of the chain, the tension of the chain at the lowest point is ________.

• A. \( \sqrt{3}mg \)
• B. \( \dfrac{\sqrt{3}}{2}mg \)
• C. \( mg \)
• D. \( \dfrac{1}{2}mg \)

Hint:

For hanging chains, tension at the lowest point equals the horizontal component of tension at supports.

Answer 1

The Correct Option is B

Step 1: Consider equilibrium of half the chain.
At the lowest point, tension is purely horizontal and is the same throughout the chain horizontally.
Step 2: Resolve tension at support.
Let tension at the support be \(T\). Vertical component balances half the weight: \[ T \sin 30^\circ = \frac{mg}{2} \]
Step 3: Solve for tension.
\[ T \times \frac{1}{2} = \frac{mg}{2} \Rightarrow T = mg \]
Step 4: Find horizontal component (tension at lowest point).
\[ T_0 = T \cos 30^\circ = mg \times \frac{\sqrt{3}}{2} \]