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A simple pendulum of length $L$ swings in a vertical plane. The tension of the string when it makes an angle $\theta$ with the vertical and the bob of mass $m$ moves with a speed $v$ is $\mathrm{fg}$ is the gravitational acceleration)
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The correct answer is:
$m g \cos \theta+m v^{2} / L$
The situation is given below
For motion along a vertical circular track, the required centripetal force is along the radius and towards the centre of the circle is given by $T-m g \cos \theta=\frac{m v^{2}}{L}$
$\Rightarrow \quad T=\frac{m v^{2}}{L}+m g \cos \theta$
For motion along a vertical circular track, the required centripetal force is along the radius and towards the centre of the circle is given by $T-m g \cos \theta=\frac{m v^{2}}{L}$
$\Rightarrow \quad T=\frac{m v^{2}}{L}+m g \cos \theta$
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