A thin uniform rod of mass M and length L is hinged at an end and released from rest in the horizontal position. The tension at a point located a distance L/3 from the hinge point, when the rod becomes vertical, will be

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$\frac{22 Mg}{27}$

$\frac{11 Mg}{13}$

$\frac{6 Mg}{11}$

2Mg

answer is D.

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Detailed Solution

When the rod becomes vertical, apply conservation of energy
$\begin{array}{l} \frac{MgL}{2} = \frac{1}{2} \frac{{ML}^{2}}{3} w^{2} \\ \Rightarrow W = \sqrt{3 g / L} \end{array}$
Centre of mass of lower two third part moves in a circle of radius 2L/3.