Laws of Motion

JEE Main level, single correct, medium

73 of 210

Question 73 of 210

A flexible chain of mass mm 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∘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: (A) 32mg\frac{\sqrt{3}}{2} \mathrm{mg}

A chain of mass mm hangs between two fixed points at the same level with inclination 30°30° at support points.
Consider equilibrium of one half of the chain (mass m/2m/2):
Horizontal equilibrium: T0=T1cos⁡(30°)=T1⋅32T_0 = T_1\cos(30°) = T_1 \cdot \frac{\sqrt{3}}{2}
Vertical equilibrium: T1sin⁡(30°)=m2gT_1\sin(30°) = \frac{m}{2}g
From vertical equation: T1⋅12=mg2T_1 \cdot \frac{1}{2} = \frac{mg}{2}, giving T1=mgT_1 = mg
Substituting into horizontal equation: T0=mg⋅32=32mgT_0 = mg \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}mg

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