$\sin \theta_0=\frac{A}{l}=\frac{10}{100}=\frac{1}{10}$

From conservation of energy

$\frac{1}{2} m v^2=m g l(1-\cos \theta)$

Maximum tension occurs at mean position.

$\therefore T-m g=\frac{m v^2}{l}$

$\Rightarrow T=m g+\frac{m v^2}{l}$

$\therefore T=m g+2 m g(1-\cos \theta)$

$=m g\left[1+2\left(1-\sqrt{1-\sin ^2 \theta}\right)\right]$

$=m g\left[3-2 \sqrt{1-\frac{1}{100}}\right]$

$=\frac{250}{1000} \times 9.8\left[3-2\left(1-\frac{1}{200}\right)\right]=\frac{99}{40}$

$\therefore x=99$