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

The Correct Option is B

Resistivity is calculated using the formula:

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

The formula for the relative error in resistivity is:

\(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \frac{\Delta r}{r} + \frac{\Delta l}{l}\)
Upon substituting the provided values:

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

The individual terms are evaluated as follows:

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

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