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

To determine the percentage error in measuring the resistance of a wire, we start by understanding the relationship between the resistance, length, and diameter of the wire. The resistance \( R \) of a wire is given by the formula:

\(R = \frac{\rho L}{A}\) 

where \(\rho\) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area of the wire. For a wire with a circular cross-section, the area \( A \) can be expressed in terms of the diameter \( d \) as:

\(A = \frac{\pi d^2}{4}\)

Substituting for \( A \) in the resistance formula gives:

\(R = \frac{4\rho L}{\pi d^2}\)

To find the percentage error in \( R \), we use the formula for percentage error propagation. For a function \( R = \frac{4 L}{d^2} \), the percentage error is given by:

\(\text{Percentage error in } R = \left( \text{percentage error in } L + 2 \times \text{percentage error in } d \right)\)

Given that the percentage errors in measuring the length \( L \) and diameter \( d \) are both \( 0.1\% \), the percentage error in resistance \( R \) becomes:

\(\text{Percentage error in } R = 0.1\% + 2 \times 0.1\%\)

Calculating this gives:

\(\text{Percentage error in } R = 0.1\% + 0.2\% = 0.3\%\)

Thus, the percentage error in measuring the resistance of the wire is 0.3\%. Therefore, the correct answer is \(0.003\) when converted to a decimal (since \( 0.3\% = 0.003 \)).

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:

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