JEE Main 2026 Jan 24 Shift 2, Physics Q15: Newton's Laws Of Motion

JEE Main2026Jan 24, Shift 2Physics

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
$\frac{3}{2} m g$
B
$\frac{1}{2} m g$
C
$m g$
D
$3 m g$

Solution

Consider one half of the chain (mass $m /2$ ). Three forces act on it: tension $T$ at the upper end (along the chain, at 30° above horizontal), tension $T_{0}$ at the lowest point (horizontal, by symmetry), and weight $\frac{m}{2} g$ downward.

Vertical equilibrium: $T sin 3 0^{\circ} = \frac{m}{2} g \Rightarrow T = m g$ .

Horizontal equilibrium: $T cos 3 0^{\circ} = T_{0} \Rightarrow T_{0} = m g \cdot \frac{3}{2} = \frac{3}{2} m g$ .

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JEE Main 2026 Jan 24 Shift 2, Physics Q15: Newton's Laws Of Motion

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