A simple pendulum with length $100 \mathrm{~cm}$ and bob of mass $250 \mathrm{~g}$ is executing S.H.M. of amplitude $10 \mathrm{~cm}$. The maximum tension in the string is found to be $\frac{x}{40} \mathrm{~N}$. The value of $x$ is ___________.

<p>Given the amplitude $A$ and the length $l$ of the pendulum, we can find the maximum angular displacement $\theta_0$:</p> <p>$\sin \theta_0 = \frac{A}{l} = \frac{10}{100} = \frac{1}{10}$</p> <p>By conservation of energy, the following equation holds:</p> <p>$\frac{1}{2} m v^2 = m g l(1 - \cos \theta)$</p> <p>The maximum tension occurs at the mean position (i.e., when the pendulum is vertical). At this point, we have:</p> <p>$$ \begin{aligned} &amp; T - mg = \frac{m v^2}{l} \ &amp; \Rightarrow T = mg + \frac{m v^2}{l} \end{aligned} $$</p> <p>Substituting the conservation of energy equation, we get:</p> <p>$$ \begin{aligned} &amp; T = mg + 2 m g(1 - \cos \theta) \\\\ &amp; = mg\left[1 + 2\left(1 - \sqrt{1 - \sin^2 \theta}\right)\right] \\\\ &amp; = mg\left[3 - 2 \sqrt{1 - \frac{1}{100}}\right] \\\\ &amp; = \frac{250}{1000} \times 10\left[3 - 2\left(1 - \frac{1}{200}\right)\right] = \frac{101}{40} \\\\ &amp; \therefore x = 101 \end{aligned} $$</p> <p>So, the value of $x$ is 101.</p>