A cylindrical wire of mass$(0.4 \pm 0.01) \mathrm{g}$ has length $(8 \pm 0.04) \mathrm{cm}$ and radius $(6 \pm 0.03) \mathrm{mm}$. The maximum error in its density will be

The formula to calculate the density of the material of the wire is given by

$\mathrm{\rho }=\frac{\mathrm{M}}{{\mathrm{\pi R}}^2\mathrm{L}} ...\left(1\right)$

Hence, the formula to calculate the percentage error in calculating the density of the material can be written as

$\frac{d\rho }{\rho }\times 100=\left[\frac{dM}{M}+\frac{2dR}{R}+\frac{dL}{L}\right]\times 100 ...\left(2\right)$

Substitute the values of the known parameters into equation (2) to calculate the required percentage error.

$\begin{array}{rcl}& =& \left[\frac{0.01}{0.4}+\frac{2\times 0.03}{6}+\frac{0.04}{8}\right]\times 100\\ & =& 4\%\end{array}$