In an experiment to find acceleration due to gravity (g) using...

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

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

Fractional error in $g$ is stated as

$\dfrac{\Deltag}{g} = \dfrac{2\DeltaT}{T} + \dfrac{\Deltal}{l}$

$⇒ \dfrac{\Deltag}{g} \times 100 = 2 \times \dfrac{1}{100 \times 0 . 5} \times 100 + \dfrac{1mm}{10cm} \times 100$

$⇒ \dfrac{\Deltag}{g} \times 100 = \dfrac{5}{100} \times 100 = 5$

Hence, value of $x = 5$.