A simple pendulum of length ' $l$ ' carries a bob of mass ' $m$ '. If the breaking strength of the string is $2 m g$ . The maximum angular amplitude from the vertical can be:

0 °

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30 °

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60 °

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90 °

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Solution

The correct option is C $60 °$
$T_{m a x} = 2 m g .$
Tension is max. at bottom
$T = m g + \frac{m v^{2}}{l}$
$2 m g = m g + \frac{m v^{2}}{l}$
Then, $v = \sqrt{l g} .$
Now by WET.
$m g l (1 - cos θ) = \frac{m}{2} (\sqrt{l g})^{2}$
$g l - g l cos θ = \frac{l g}{2}$
$\frac{g l}{2} = g l cos θ$
$cos θ = \frac{1}{2}$
$θ = 60^{0}$

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