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

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$	heta = \cos^{-1}(2/3)$

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(C) Let the mass of the bob be $m$ and the length of the string be $r$. The breaking strength of the string is given as $T_{max} = 2mg$.When the bob is at an angle $	heta$ with the vertical,the forces acting along the radial direction are the tension $T$ and the component of gravity $mg \cos 	heta$.The net centripetal force is given by:$T - mg \cos 	heta = \frac{mv^2}{r} \implies T = mg \cos 	heta + \frac{mv^2}{r}$Using the principle of conservation of energy between the horizontal position (point $A$) and the position at angle $	heta$ (point $B$):Loss in potential energy = Gain in kinetic energy$mg(r - r \cos 	heta) = \frac{1}{2}mv^2 \implies v^2 = 2gr(1 - \cos 	heta)$Substituting $v^2$ into the tension equation:$T = mg \cos 	heta + \frac{m}{r} [2gr(1 - \cos 	heta)]$$T = mg \cos 	heta + 2mg - 2mg \cos 	heta = 2mg - mg \cos 	heta$At the point of breaking,$T = 2mg$:$2mg = 2mg - mg \cos 	heta$This implies $\cos 	heta = 0$,so $	heta = 90^\circ$. However,checking the provided solution logic in the prompt: $T = mg \cos 	heta + \frac{mv^2}{r} = 2mg$. If the bob starts from horizontal,the velocity at angle $	heta$ is $v^2 = 2gr \cos 	heta$ relative to the lowest point? No,the potential energy drop from horizontal is $h = r \cos 	heta$. Thus $v^2 = 2g(r \cos 	heta)$.Substituting this: $T = mg \cos 	heta + \frac{m}{r}(2gr \cos 	heta) = 3mg \cos 	heta$.Setting $T = 2mg$: $3mg \cos 	heta = 2mg \implies \cos 	heta = 2/3$.Thus,$	heta = \cos^{-1}(2/3)$.

The breaking strength of the string of a simple pendulum is twice the weight of the bob. The bob is released from rest when the string is horizontal. At what angle $\theta$ with the vertical will the string break?

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