Question

The radius ($r$), length ($l$) and resistance ($R$) of a thin wire are $(0.20\pm 0.01)$ cm, $(800\pm 0.1)$ cm, and $(301\pm 1)\Omega$ respectively. The error in the specific resistance is :

• A. 10.34%
• B. 45.4%
• C. 26%
• D. 27.5%

Answer 1

Answer: (A)

10.34%

Here, $R=\frac{\rho L}{\pi r^2}$ where $\rho =$ specific resistance or resistivity.

or $\rho =\frac{\pi r^{2}R}{L}$

so, $\frac{\Delta \rho }{\rho }=2\frac{\Delta r}{r}+\frac{\Delta R}{R}+\frac{\Delta L}{L}=2\times \frac{0.01}{0.20}+\frac{1}{301}+\frac{0.1}{800}=0.1034$

% error $=0.1034\times 100=10.34\%$