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:
The percentage error in the resistivity of the material of the wire is:
- \( 25.6\% \)
- \( 39.9\% \)
- \( 37.3\% \)
- \( 35.6\% \)
The Correct Option is B
Solution and Explanation
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:
\(\frac{\Delta \rho}{\rho} = 0.1 + 2 \times 0.1429 + 0.0133\)
\(\frac{\Delta \rho}{\rho} \approx 0.1 + 0.2858 + 0.0133 = 0.3991 \approx 39.9\%\)
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