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(C) Let the mass of the bob be $m$ and the length of the string be $\ell$. The breaking strength of the string is given as $T_{max} = 2mg$.When the bo

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$	heta = \cos^{-1}\left(\frac{2}{3}ight)$

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(C) Let the mass of the bob be $m$ and the length of the string be $\ell$. The breaking strength of the string is given as $T_{max} = 2mg$.When the bob is released from the horizontal position,its velocity $v$ at an angle $	heta$ with the vertical is given by the conservation of energy: $\frac{1}{2}mv^2 = mg\ell \cos 	heta$,which implies $v^2 = 2g\ell \cos 	heta$.The tension $T$ in the string at any angle $	heta$ is given by $T = mg \cos 	heta + \frac{mv^2}{\ell}$.Substituting the expression for $v^2$,we get $T = mg \cos 	heta + \frac{m(2g\ell \cos 	heta)}{\ell} = mg \cos 	heta + 2mg \cos 	heta = 3mg \cos 	heta$.The string breaks when $T = T_{max} = 2mg$.Therefore,$3mg \cos 	heta = 2mg$,which simplifies to $\cos 	heta = \frac{2}{3}$.Thus,$	heta = \cos^{-1}\left(\frac{2}{3}ight)$.

In a simple pendulum,the breaking strength of the string is double the weight of the bob. The bob is released from rest when the string is horizontal. The string breaks when it makes an angle $\theta$ with the vertical.

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