Question

For a conducting wire, its resistance, $R=65\pm 1\Omega$, length, $l=5\pm 0.1mm$ and diameter, $d=10\pm 0.5mm$. Find error in calculation of its resistivity.

• A. $21\%$
• B. $13\%$
• C. $16\%$
• D. $41\%$

Answer 1

Answer: (B)

Given, $R=65\Omega ,\Delta R=1\Omega ,l=5mm=5\times 10^{-3}m$, $\Delta l=0.1mm=0.1\times 10^{-3}m,d=10mm=10\times 10^{-3}m$, $\Delta d=0.5mm=0.5\times 10^{-3}m$
$\therefore$ Resistivity, $\rho =\frac{RA}{l}$
$\Rightarrow \rho =\frac{R\pi (d/2{)}^2}{l}=\frac{\pi R{d}^2}{4l}$
$\therefore \frac{\Delta \rho }{\rho }=\frac{\Delta R}{R}+2\frac{\Delta d}{d}+\frac{\Delta l}{l}$
Substituting the given values, we get
$\frac{\Delta \rho }{\rho }=\frac{1}{65}+2\left(\frac{0.5\times 10^{-3}}{10\times 10^{-3}}\right)+\frac{0.1\times 10^{-3}}{5\times 10^{-3}}$
$\Rightarrow \frac{\Delta \rho }{\rho }=0.0154+0.1+0.02$
$\Rightarrow \frac{\Delta \rho }{\rho }=0.1354$
So, $\%$ error in calculation of resistivity $=0.1354\times 100\%$
$=13.5\%\approx 13\%$.