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

Given that:-

length $=l_{m}$.

mass of bob $=m \mathrm{~kg}$.

Breaking strength $=2 \mathrm{mg}$.

We know that Maximum tension is at the bottom

$\begin{array}{l} \Rightarrow T=m g+\frac{m v^{2}}{l} \\ \Rightarrow 2 m g=m g+\frac{m v^{2}}{l} \\ \Rightarrow v=\sqrt{g l} \end{array}$

Now by work-energy theorem:-

$\begin{array}{l} \Rightarrow m g h=\frac{m v^{2}}{2} \\ \Rightarrow m g l(1-\cos \theta)=\frac{m}{2}(\sqrt{l g})^{2} \\ \Rightarrow g l-g l \cos \theta=\frac{g l}{2} \\ \Rightarrow \frac{g l}{2}=g l \cos \theta \\ \Rightarrow \cos \theta=\frac{1}{2} \\ \Rightarrow \theta=60^{\circ} \end{array}$

Hence, option C is the Correct answer.

Given that:-

length $=l_{m}$.

mass of bob $=m \mathrm{~kg}$.

Breaking strength $=2 \mathrm{mg}$.

We know that Maximum tension is at the bottom

$\begin{array}{l} \Rightarrow T=m g+\frac{m v^{2}}{l} \\ \Rightarrow 2 m g=m g+\frac{m v^{2}}{l} \\ \Rightarrow v=\sqrt{g l} \end{array}$

Now by work-energy theorem:-

$\begin{array}{l} \Rightarrow m g h=\frac{m v^{2}}{2} \\ \Rightarrow m g l(1-\cos \theta)=\frac{m}{2}(\sqrt{l g})^{2} \\ \Rightarrow g l-g l \cos \theta=\frac{g l}{2} \\ \Rightarrow \frac{g l}{2}=g l \cos \theta \\ \Rightarrow \cos \theta=\frac{1}{2} \\ \Rightarrow \theta=60^{\circ} \end{array}$

Hence, option C is the Correct answer.