Length of a simple pendulum (100.0±0.5) cm, and time period T = (2.00 ± 0.01) s. Determine the percentage of error in acceleration due to gravity 'g'.

The acceleration due to gravity $g$ can be determined from the formula for the time period $T$ of a simple pendulum:

$T=2 \pi \sqrt{\frac{L}{g}}$

$T^{2}=4 \pi^{2} \frac{L}{g}$

$g=\frac{4 \pi^{2} L}{T^{2}}$

Now, we are asked to determine the percentage error in $g$. The general formula for the percentage error in a quantity derived from other quantities is given by:

$\text { Percentage Error in } g=\left(\frac{\Delta g}{g}\right) \times 100$

Using propagation of errors, for $g=\frac{4 \pi^{2} L}{T^{2}}$, the relative errors add as follows:

$\frac{\Delta g}{g}=\frac{\Delta L}{L}+2 \times \frac{\Delta T}{T}$

Where:

- $\Delta L=0.5 \mathrm{~cm}=0.005 \mathrm{~m}$

- $L=100.0 \mathrm{~cm}=1.00 \mathrm{~m}$

- $\Delta T=0.01 \mathrm{~s}$

- $T=2.00 \mathrm{~s}$

Now, substitute the values into the formula:

$\begin{array}{c} \frac{\Delta g}{g}=\frac{0.005}{1.00}+2 \times \frac{0.01}{2.00} \\ \frac{\Delta g}{g}=0.005+2 \times 0.005=0.005+0.01=0.015 \end{array}$

Therefore, the percentage error in $g$ is:

Percentage Error in $g=0.015 \times 100=1.5 \%$

Thus, the percentage error in the acceleration due to gravity $g$ is $1.5 \%$.