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°. Considering the equilibrium of each half of the chain, the tension of the chain at the lowest point is ______. (a) mg (b) √(3)/2mg (c) 1/2mg (d) √(3)mg

Answer: (b). Take one half of the chain (weight mg/2). It is pulled by T at the support ( 30° to the horizontal) and by T₀ (horizontal) at the lowest point. Vertical: Tsin 30° = mg/2 ⇒ T = mg Horizontal: T₀ = Tcos 30° = √(3)/2mg So the answer is (b) √(3)/2mg.

A flexible chain of mass m hangs between two fixed points at the same… | JEE Main 2026 (24 Jan, Shift 2)

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Equilibrium, Third Law & Frames

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 ______.

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Answer: (b)

Take one half of the chain (weight $\frac{mg}{2}$ ). It is pulled by $T$ at the support ( $30^\circ$ to the horizontal) and by $T_0$ (horizontal) at the lowest point.
Vertical: $T\sin 30^\circ = \frac{mg}{2} \Rightarrow T = mg$
Horizontal: $T_0 = T\cos 30^\circ = \frac{\sqrt{3}}{2}mg$
So the answer is (b) $\frac{\sqrt{3}}{2}mg$ .

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