Physics - Error Question with Solution | TestHub

Given: $\begin{array}{rl}g& =\frac{4{\pi }^2\mathrm{L}}{{\mathrm{T}}^2}\\ \mathrm{L}& =(10\pm 0.1)\mathrm{cm},\text{ and }\\ \mathrm{T}& =(100\pm 1)\mathrm{s}\end{array}$ Hence, error in measurement of $g$ is $\begin{array}{rl}\frac{\Delta g}{g}& =\frac{\Delta \mathrm{L}}{\mathrm{L}}+2\frac{\Delta \mathrm{T}}{\mathrm{T}}\\ & =\frac{0.1}{10}\times 100+2\times \frac{1}{100}\times 100\\ & =3\%\end{array}$