Find the percentage error in calculating the volume of a cubical box, if an error of $1\%$ is made in measuring the length of edges of the cube.
Let the length of edge of cube $= \ell$
Given: $\dfrac{\Delta \ell}{\ell} = 1 \% = 0.01$
Now, volume of cube $= V = \ell^3$
$\therefore$ $\dfrac{dV}{d\ell} = 3\ell^2$
Now, $\Delta V = \left(\dfrac{dV}{d\ell}\right) \times \Delta \ell$
$\begin{aligned}
\% \text{Error in V} = \dfrac{\Delta V}{V} \times 100 \% & = \dfrac{\left(dV/d \ell\right) \times \Delta \ell}{V} \times 100 \% \\
& = \dfrac{3 \ell^2 \times \Delta \ell}{\ell^3} \times 100 \% \\
& = 3 \times \left(\dfrac{\Delta \ell}{\ell}\right) \times 100 \% \\
& = 3 \times 0.01 \times 100 \% \\
& = 3 \%
\end{aligned}$