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

The formula for the acceleration due to gravity is $g = \dfrac{4\pi^2 L}{T^2}$

Since the time for $n$ oscillations is $t = nT$, we have $T = \dfrac{t}{n}$. Thus, the relative error in $T is the same as the relative error in$t$:$\dfrac{\Delta T}{T} = \dfrac{\Delta t}{t}$.<br><br>The expression for the maximum percentage error in$g$is:<br>\left(\dfrac{\Delta g}{g} \times 100\right) = \left(\dfrac{\Delta L}{L} \times 100\right) + 2\left(\dfrac{\Delta t}{t} \times 100\right)$

From the given data:
$L = 100.0 \text{ cm}$, $\Delta L = 1 \text{ mm} = 0.1 \text{ cm}$
$t = 40.0 \text{ s}$, $\Delta t = 0.1 \text{ s}$

Calculating the individual percentage errors:
Percentage error in $L = \dfrac{0.1}{100.0} \times 100 = 0.1 \%$
Percentage error in $t = \dfrac{0.1}{40.0} \times 100 = 0.25 \%$

Substituting these into the error equation:
Maximum percentage error in $g = 0.1 \% + 2(0.25 \%) = 0.1 \% + 0.5 \% = 0.6 \%$

Answer: $0.6 \%$