Question

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

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

Answer 1

Answer: (B)

Given, $R=65\, \Omega, \Delta R=1 \, \Omega, l=5\, mm =5 \times 10^{-3} m$, $\Delta l=0.1 \, mm =0.1 \times 10^{-3} m , d=10\, mm =10 \times 10^{-3} m$, $\Delta d=0.5 \, mm =0.5 \times 10^{-3} m$
$\therefore$ Resistivity, $\rho=\frac{R A}{l}$
$ \Rightarrow \rho=\frac{R \pi(d / 2)^{2}}{l}=\frac{\pi R d^{2}}{4 l}$
$\therefore \frac{\Delta \rho}{\rho}=\frac{\Delta R}{R}+2 \frac{\Delta d}{d}+\frac{\Delta l}{l}$
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 $=0.1354 \times 100 \%$
$=13.5 \% \approx 13 \%$.