JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Laws of Motion

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

3 0^{\circ}

. Considering the equilibrium of each half of the chain, the tension of the chain at the lowest point is ____.

Answer: (A)

Solution

Take one half of the chain in equilibrium. Its weight

\frac{m g}{2}

acts downward, the support tension

T

acts along the chain at

3 0^{\circ}

to the horizontal, and the tension

T_{0}

at the lowest point acts horizontally. Vertical balance gives

T sin 3 0^{\circ} = \frac{m g}{2}

and horizontal balance gives

T cos 3 0^{\circ} = T_{0}

. Dividing,

T_{0} = \frac{m g}{2} cot 3 0^{\circ} = \frac{m g}{2} 3

. The tension at the lowest point is

\frac{3}{2} m g

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Single correct option