Question

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

• A. 13%
• B. 21%
• C. 16%
• D. 41%

Answer 1

Answer: (A)

$R=\frac{\rho l}{\pi\left(d / 2\right)^{2}} \Rightarrow R=\frac{4\rho l}{\pi d^{2}} \Rightarrow \rho=\frac{\pi Rd^{2}}{4l}$
$\frac{\Delta\rho}{\rho}=\frac{\Delta R}{R}+\frac{2\Delta d}{d}+\frac{\Delta l}{l}$
$\Rightarrow \frac{\Delta\rho}{\rho}=\frac{1}{65}+\frac{2.\left(0.5\right)}{10}+\frac{0.1}{5}$
$\Rightarrow \frac{\Delta\rho}{\rho}=0.015+0.1+0.02$
$\Rightarrow \frac{\Delta\rho}{\rho}\approx0.13$
So 13% error.