Question

The radius ($r$), length ($l$), and resistance ($R$) of a metal wire were measured in the laboratory as: $r = (0.35 \pm 0.05) \, \text{cm}, \quad R = (100 \pm 10) \, \text{ohm}, \quad l = (15 \pm 0.2) \, \text{cm}.$
The percentage error in the resistivity of the material of the wire is:

• A. $25.6\%$
• B. $39.9\%$
• C. $37.3\%$
• D. $35.6\%$

Answer 1

Answer: (B)

$39.9\%$

Answer 2

The formula for resistivity is given by:

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

The relative error in resistivity is given by:

$\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \frac{\Delta r}{r} + \frac{\Delta l}{l}$
Substituting the given values:

$\frac{\Delta \rho}{\rho} = \frac{10}{100} + 2 \times \frac{0.05}{0.35} + \frac{0.2}{15}$

Calculating each term:

$\frac{\Delta \rho}{\rho} = 0.1 + 2 \times 0.1429 + 0.0133$

$\frac{\Delta \rho}{\rho} \approx 0.1 + 0.2858 + 0.0133 = 0.3991 \approx 39.9\%$