Question

Find error in calculation of resistivity of
$R=65\pm 1\Omega$;
$I=5\pm 0.1mm$
$d=10\pm 0.5mm$

• A. 13%
• B. 21%
• C. 16%
• D. 41%

Answer 1

Answer: (A)

$R=\frac{\rho l}{\pi {\left(d/2\right)}^2}\Rightarrow R=\frac{4\rho l}{\pi d^2}\Rightarrow \rho =\frac{\pi R{d}^2}{4l}$
$\frac{\Delta \rho }{\rho }=\frac{\Delta R}{R}+\frac{2\Delta d}{d}+\frac{\Delta l}{l}$
$\Rightarrow \frac{\Delta \rho }{\rho }=\frac{1}{65}+\frac{2.\left(0.5\right)}{10}+\frac{0.1}{5}$
$\Rightarrow \frac{\Delta \rho }{\rho }=0.015+0.1+0.02$
$\Rightarrow \frac{\Delta \rho }{\rho }\approx 0.13$
So 13% error.