Question

Calculate the percentage error in specific resistance, $\rho =\pi r^{2}R/l$, where $r=$ radius of wire $=0.26\pm 0.02cm,l=$ length of wire $=156.0\pm 0.1cm$, and $R=$ resistance of wire $=64\pm 2\Omega$.

Answer 1

$P=\frac{{\Pi r}^{2}R}{l}$

$\frac{\Delta P}{P}=\frac{2\Delta r}{r}+\frac{\Delta R}{R}+\frac{\Delta l}{l}$

$=2\times \frac{0.02}{0.26}+\frac{2}{64}+\frac{0.1}{156}$

$=0.186$

$\therefore$ % error $=18.6$%