Question

The specific resistance $\rho$ of a circular wire of radius r, resistance R, and length l is given by $\rho =\pi r^{2}R/l$. Given: $r=0.24\pm 0.02cm,R=30\pm 1\Omega ,$ and $l=4.80\pm 0.01cm.$ The percentage error in $\rho$ is nearly:

• A. $7\%$
• B. $9\%$
• C. $13\%$
• D. $20\%$

Answer 1

Answer: (D)

$20\%$

The specific resistance is given as :

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

Differentiating both sides given (in terms of relative error ):-

$\frac{\Delta P}{\rho }=2\frac{\Delta r}{r}+\frac{\Delta R}{R}+\frac{\Delta l}{l}\to \left(2\right)$

From given data,

$r=0.24\pm 0.02cm\Rightarrow \frac{\Delta r}{r}=\frac{0.02}{0.24}=\frac{1}{12}$

$R=30\pm 1\Omega \Rightarrow \frac{\Delta R}{R}=\frac{1}{30}$

$l=4.80\pm 0.01cm\Rightarrow \frac{\Delta L}{L}=\frac{0.07}{4.80}=\frac{1}{480}$

Hence from (1),

$\frac{\Delta P}{\rho }=2\times \frac{1}{12}+\frac{1}{30}+\frac{1}{480}=20.2\%$