A student measures the effective length of a simple pendulum as 100.0 ± 0.1 cm…

Step 1. The period and its error. $T = \dfrac{40.2\ \text{s}}{20} = 2.01$ s. $\Delta T = \dfrac{0.1\ \text{s}}{20} = 0.005$ s.

Step 2. The value of g. $g = \dfrac{4\pi^2 L}{T^2}$ $= \dfrac{4\pi^2 \times 1.000\ \text{m}}{(2.01\ \text{s})^2}$ $= 9.77\ \text{m/s}^2$.

Step 3. The relative error. $\dfrac{\Delta g}{g} = \dfrac{0.001}{1.000} + 2 \times \dfrac{0.005}{2.01}$ $= 0.0010 + 0.0050 = 0.0060$. That is $0.60\%$.

Step 4. The absolute error. $\Delta g = 0.0060 \times 9.77\ \text{m/s}^2$ $= 0.06\ \text{m/s}^2$.