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A flexible chain of mass m hangs between two fixed points A and B at the same level. The inclination of the chain with the horizontal at the two points of the support is $θ$ . The tension at the mid point of the chain is

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\frac{mg}{\tan θ}

\frac{mg}{2 \tan θ}

zero

\frac{mg (\sin θ + \cos θ)}{2 \tan θ}

answer is B.

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Detailed Solution

let $T^{1}$ is the Tension at end

T is the Tension at mid point

$T^{1} \sin θ = \frac{mg}{2} \to (1)$

$T^{1} \cos θ = T \to (2)$

$Tan θ = \frac{mg}{2T}$

$T = \frac{mg}{2 Tan θ} = \frac{mg \cot θ}{2}$

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