Question

Find the percentage error in specific resistance given by $\rho =\frac{\pi r^{2}R}{l}$ where r is the radius having value $(0.2\pm 0.02)$ cm, R is the resistance of $(60\pm 2)\Omega$ and l is length of $(150\pm 0.1)$ cm.

• A. 5.85%
• B. 11.7%
• C. 23.4%
• D. 35.1%

Answer 1

Answer: (C)

23.4%

Applying logarithm on both sides of the given expression and differentiating,

we get $\frac{\Delta \rho }{\rho }=\pm (2\frac{\Delta r}{r}+\frac{\Delta l}{l}+\frac{\Delta R}{R})$

Given : $\Delta r$=0.02cm, $\Delta R$=2 ohm, $\Delta l$=0.1cm

Substituting the values in above expression,

$\frac{\Delta \rho }{\rho }=\pm (2\frac{0.02}{0.2}+\frac{0.1}{150}+\frac{2}{60})=0.234=23.4$%