The time period of a simple pendulum is given as $T = 2\pi \sqrt{\dfrac{\ell}{g}}$. If there is a $4\%$ error in measuring the period of the pendulum, find the approximate percentage error in the measurement of length $\ell$ of the pendulum. g is the acceleration due to gravity, a constant.
$\dfrac{\Delta T}{T} = 4\% = \dfrac{4}{100}$
$T = 2\pi \sqrt{\dfrac{\ell}{g}}$
$\therefore$ $\log T = \log \left(2\pi\right) + \dfrac{1}{2} \log \left(\ell\right) - \dfrac{1}{2} \log \left(g\right)$
$\therefore$ $\dfrac{1}{T} \times \dfrac{dT}{d\ell} = \dfrac{1}{2} \times \dfrac{1}{\ell}$
i.e. $\dfrac{dT}{d\ell} = \dfrac{T}{2 \ell} \;\; \cdots$ (1)
Now, $\Delta T = \left(\dfrac{dT}{d\ell}\right) \times \Delta \ell$
$\implies$ $\Delta \ell = \dfrac{\Delta T}{\left(dT / d\ell\right)}$
$\begin{aligned}
\therefore \% \text{Error in } \ell = \dfrac{\Delta \ell}{\ell} \times 100 \% & = \dfrac{\Delta T}{\ell \times \left(dT / d\ell\right)} \times 100 \% \\
& = \dfrac{\Delta T \times 2 \ell}{\ell \times T} \times 100 \% \;\; \left[\text{From equation (1)}\right] \\
& = 2 \times \dfrac{\Delta T}{T} \times 100 \% \\
& = 2 \times \dfrac{4}{100} \times 100 = 8 \%
\end{aligned}$