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(C) Let the maximum angular amplitude be $	heta$. When the pendulum bob moves from the extreme position $B$ to the lowest point $A$,the decrease in p

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$60$

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(C) Let the maximum angular amplitude be $	heta$. When the pendulum bob moves from the extreme position $B$ to the lowest point $A$,the decrease in potential energy equals the increase in kinetic energy at $A$.The vertical height difference between $B$ and $A$ is $h = l - l \cos 	heta = l(1 - \cos 	heta)$.By the law of conservation of energy:$mgh = \frac{1}{2}mv^2$$mg(l(1 - \cos 	heta)) = \frac{1}{2}mv^2$$v^2 = 2gl(1 - \cos 	heta)$At the lowest point $A$,the forces acting on the bob are the tension $T$ upwards and weight $mg$ downwards. The net centripetal force is:$T - mg = \frac{mv^2}{l}$Substituting $v^2$:$T - mg = \frac{m}{l} \cdot 2gl(1 - \cos 	heta)$$T - mg = 2mg(1 - \cos 	heta)$$T = mg + 2mg - 2mg \cos 	heta = mg(3 - 2 \cos 	heta)$The string breaks when $T = T_{\max} = 2mg$. Setting $T = 2mg$:$2mg = mg(3 - 2 \cos 	heta)$$2 = 3 - 2 \cos 	heta$$2 \cos 	heta = 1$$\cos 	heta = \frac{1}{2}$$	heta = 60^\circ$

$A$ pendulum consisting of a small sphere of mass $m$ suspended by an inextensible and massless string of length $l$ is made to swing in a vertical plane. If the breaking strength of the string is $2mg$,then the maximum angular amplitude of the displacement from the vertical can be ....... $^o$

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