A wire has a mass 0.3 ± 0.003 g radius . 0.5 ± 0.005 mm and length 6 ± 0.06 cm. The maximum percentage error in the measurement of its density is

The correct option is 4.

To find the maximum percentage error in the measurement of the density of the wire, we start with the formula for the density ($\rho$) of a cylindrical wire:
$\rho =\frac{\text{Mass}}{\text{Volume}}=\frac{m}{\pi r^{2}l}$
where:
$m$ is the mass of the wire,
$r$ is the radius of the wire, and
$l$ is the length of the wire.

The relative error in density, $\frac{\Delta \rho }{\rho }$, is given by the sum of the relative errors of the measured quantities:
$\frac{\Delta \rho }{\rho }=\frac{\Delta m}{m}+2\frac{\Delta r}{r}+\frac{\Delta l}{l}$

The maximum percentage error in density is:
$\frac{\Delta \rho }{\rho }\times 100=(\frac{\Delta m}{m}+2\frac{\Delta r}{r}+\frac{\Delta l}{l})\times 100$

Given values from the problem:
Mass, $m=0.3\pm 0.003\text{g}$ $\Rightarrow \Delta m=0.003$ and $m=0.3$
Radius, $r=0.5\pm 0.005\text{mm}$ $\Rightarrow \Delta r=0.005$ and $r=0.5$
Length, $l=6\pm 0.06\text{cm}$ $\Rightarrow \Delta l=0.06$ and $l=6$

Substituting these values into the percentage error formula:
$\frac{\Delta \rho }{\rho }\times 100=(\frac{0.003}{0.3}+2\times \frac{0.005}{0.5}+\frac{0.06}{6})\times 100$

Calculate each term:
$\frac{\Delta m}{m}=\frac{0.003}{0.3}=0.01$
$2\frac{\Delta r}{r}=2\times \frac{0.005}{0.5}=0.02$
$\frac{\Delta l}{l}=\frac{0.06}{6}=0.01$

Add the terms together:
$\frac{\Delta \rho }{\rho }\times 100=(0.01+0.02+0.01)\times 100$
$\frac{\Delta \rho }{\rho }\times 100=0.04\times 100=4\%$

Thus, the maximum percentage error in the measurement of the density is 4%.