If a copper wire is stretched to increase its length by \(20 \%\) The percentage increase in resistance of the wire is _______\(\%\).
If a copper wire is stretched to increase its length by \(20 \%\) The percentage increase in resistance of the wire is _______\(\%\).
Show Hint
When a wire is stretched, the resistance increases as R ∝ L2 (if volume remains constant).
Correct Answer: 44
Approach Solution - 1
Solution:
The resistance of a wire is given by:
\[ R = \rho \frac{L}{A} \]
where \(L\) is the length and \(A\) is the cross-sectional area. When a wire is stretched, its volume remains constant:
\[ L_1 A_1 = L_2 A_2 \]
As \(L_2 = 1.2L_1\), the area decreases as:
\[ A_2 = \frac{A_1}{1.2} \]
The new resistance is:
\[ R_2 = \rho \frac{L_2}{A_2} = \rho \frac{1.2L_1}{A_1 / 1.2} = 1.44R_1 \]
The percentage increase in resistance is:
\[ \% \, \text{Increase} = \frac{R_2 - R_1}{R_1} \times 100 = \frac{1.44R_1 - R_1}{R_1} \times 100 = 44\% \]
Thus, the resistance increases by 44%.
Approach Solution -2
The correct answer is 44.
As volume is constant,
So resistance ∝ (length )2
⇒% change in resistance
\(\frac{1.44 R_0}{R_0}×100=0.44×100\)
\(=44\%\)
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