In an experiment to find acceleration due to gravity g using simple pendulum, time period of 0 . 5   s…

Time period of oscillation is given by $T=2\pi \sqrt{\frac{l}{g}}$ , here, $l$ is length and $g$ is acceleration due to gravity.

From above relation, we have  $g=\frac{1}{4{\pi }^2}\frac{{\mathrm{T}}^2}{l}$.

Fractional error in $g$ is stated as 

$\frac{\Delta g}{g}=\frac{2\Delta T}{T}+\frac{\Delta l}{l}$

$\Rightarrow \frac{\Delta g}{g}\times 100=2\times \frac{1}{100\times 0.5}\times 100+\frac{1 \mathrm{mm}}{10 \mathrm{cm}}\times 100$

$\Rightarrow \frac{\Delta g}{g}\times 100=\frac{5}{100}\times 100=5\%$

Hence, value of $x=5$.