Question

If the percentage errors in measuring the length and the diameter of a wire are 0.1% each. The percentage error in measuring its resistance will be:

• A. 0.002
• B. 0.003
• C. 0.001
• D. 0.00144

Answer 1

The Correct Option is B

The resistance \( R \) of the wire is given by:

\[ R = \frac{\rho L}{\pi d^2 / 4} \]

Taking the percentage error:

\[ \frac{\Delta R}{R} = \frac{\Delta L}{L} + 2 \frac{\Delta d}{d} \]

Given:

\[ \frac{\Delta L}{L} = 0.1\% \quad \text{and} \quad \frac{\Delta d}{d} = 0.1\% \]

Therefore:

\[ \frac{\Delta R}{R} = 0.1\% + 2 \times 0.1\% = 0.3\% =0.003 \]