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

The acceleration due to gravity $g$ determined by a simple pendulum is given by $g = \dfrac{4\pi^2 L}{T^2}$.

The maximum percentage error in $g$ is given by the sum of the fractional errors of the individual quantities, taking into account their powers:

$\dfrac{\Delta g}{g} \times 100 = \dfrac{\Delta L}{L} \times 100 + 2 \left( \dfrac{\Delta T}{T} \times 100 \right)$

Given $L = (100.0 \pm 0.1) \text{ cm}$ and $T = (2.00 \pm 0.02) \text{ s}$, we calculate the individual percentage errors:

Percentage error in $L = \left( \dfrac{0.1}{100.0} \right) \times 100 = 0.1\%$

Percentage error in $T = \left( \dfrac{0.02}{2.00} \right) \times 100 = 1.0\%$

Substituting these into the error equation:

Maximum percentage error in $g = 0.1\% + 2(1.0\%) = 2.1\%$

Answer: $2.1\%$