Physics · Newton's Laws of Motion

JEE Main 2026 — 24 January, Evening Shift — Question 40

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 $\_\_\_\_$ .

Option A:
$\frac{\sqrt{3}}{2} m g$
Correct
Option B:
$\frac{1}{2} m g$
Option C:
$m g$
Option D:
$\sqrt{3} \mathrm{mg}$

Answer: A

Step-by-step solution

\begin{aligned} & \mathrm{T} \sin 30^{\circ}=\frac{\mathrm{m}}{2} \mathrm{~g} & \mathrm{~T} \cos 30^{\circ}=\mathrm{T}_{0} & \tan 30^{\circ}=\frac{\mathrm{mg}}{2 \mathrm{~T}_{0}} & \mathrm{~T}_{0}=\frac{\sqrt{3}}{2} \mathrm{mg} \end{aligned}

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 24 January, Evening Shift
Subject: Physics
Chapter: Newton's Laws of Motion
Topic: Free Body Diagrams and Constraint Relations