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

Answer: (B)

0.003

Answer 2

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:

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