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:
- 0.002
- 0.003
- 0.001
- 0.00144
The Correct Option is B
Approach Solution - 1
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 \( 0.3\% = 0.003 \)).
Approach Solution -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:
\[ \frac{\Delta R}{R} = 0.1\% + 2 \times 0.1\% = 0.3\% =0.003 \]
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