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MCQEasyJEE 2026Equilibrium of Rigid Bodies

JEE Physics 2026 Question with Solution

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 _____.

  • A

    3mg\sqrt{3}mg

  • B

    32mg\dfrac{\sqrt{3}}{2}mg

  • C

    mgmg

  • D

    12mg\dfrac{1}{2}mg

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: A flexible chain of mass mm hangs symmetrically between two fixed points at the same level, and the inclination at each support is 30∘30^\circ.

Find: The tension at the lowest point of the chain.

Consider the equilibrium of one half of the chain. At the lowest point, the tension acts horizontally. Let the tension at the support be TT and the tension at the lowest point be T0T_0.

For one half of the chain, the vertical component of the support tension balances half the weight:

Tsin⁡30∘=mg2T \sin 30^\circ = \frac{mg}{2}

Using sin⁡30∘=12\sin 30^\circ = \frac{1}{2},

T×12=mg2T \times \frac{1}{2} = \frac{mg}{2}

so,

T=mgT = mg

The tension at the lowest point equals the horizontal component of the support tension:

T0=Tcos⁡30∘T_0 = T \cos 30^\circ

Substituting T=mgT = mg,

T0=mg×32T_0 = mg \times \frac{\sqrt{3}}{2}

Therefore, the tension at the lowest point is 32mg\dfrac{\sqrt{3}}{2}mg. The correct option is B.

Common mistakes

  • Using the full weight mgmg for one half of the chain is incorrect, because each half supports only half the total weight. For equilibrium of one half, use mg2\frac{mg}{2} in the vertical balance.

  • Taking the tension at the lowest point equal to the support tension TT is incorrect. The lowest-point tension is only the horizontal component of the support tension, so use T0=Tcos⁡30∘T_0 = T\cos 30^\circ.

  • Interchanging sine and cosine at the support gives the wrong result. Since the chain makes 30∘30^\circ with the horizontal, the vertical component is Tsin⁡30∘T\sin 30^\circ and the horizontal component is Tcos⁡30∘T\cos 30^\circ.

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