Question

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

Answer 1

Given;
$m\pm \Delta m=(0.3\pm 0.003)g$
$r\pm \Delta r=(0.5\pm 0.005)mm$
$l\pm \Delta l=(6\pm 0.06)cm$
As we know
$\rho =\frac{m}{v}=\frac{m}{\text{Volume of wire}}$

Volume of wire =$\pi r^{2}l$

Therefore,
$\rho =\frac{m}{\pi rl}\dots \left(i\right)$

$\left(\frac{\Delta \rho }{\rho }\right)\times 100=(\frac{\Delta m}{m}+\frac{2\Delta r}{r}+\frac{\Delta l}{l})\times 100\dots \left(ii\right)$

By putting the values in equation $\left(ii\right)$

$\left(\frac{\Delta \rho }{\rho }\right)\times 100=(\frac{0.003}{0.3}+2\times \frac{0.005}{0.5}+\frac{0.06}{6})\times 100$

$\left(\frac{\Delta \rho }{\rho }\right)\times 100=(0.01+0.02+0.01)\times 100$

$\left(\frac{\Delta \rho }{\rho }\right)\%=4\%$

Final answer : $\left(\frac{\Delta \rho }{\rho }\right)\%=4\%$