In an experiment to determine the acceleration due to gravity ( g )...

The time period of a simple pendulum is given by $T = 2\pi\sqrt{\dfrac{l}{g}}$.

Squaring both sides and rearranging for $g$, we get:
$g = \dfrac{4\pi^2 l}{T^2}$

The maximum relative error in $g$ is given by:
$\dfrac{\Delta g}{g} = \dfrac{\Delta l}{l} + 2\dfrac{\Delta T}{T}$

Percentage error in length $l$ is:
$\dfrac{\Delta l}{l} \times 100 = \dfrac{0.1}{100.0} \times 100 = 0.1\%$

Percentage error in time period $T$ is:
$\dfrac{\Delta T}{T} \times 100 = \dfrac{0.02}{2.00} \times 100 = 1.0\%$

Therefore, the maximum percentage error in $g$ is:
\left(\dfrac{\Delta g}{g} \times 100\right) = 0.1\% + 2(1.0\%) = 2.1\%$<br><br>Answer:$2.1\%