From conservation of energy

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

Maximum tension occurs at mean position.

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

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

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

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

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

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

\(\therefore x=99\)