If the length of a copper wire is increased by 5% due to stretching. What will be the percentage increase in its resistance?
Options:
A .  
12%
B .  
11.5%
C .  
20.25%
D .  
10.25%
Answer: Option D : D Resistance, R=ρLA where: ρ: Resistivity, L: Length of the wire A: Cross-sectional area of the wire. The length of the copper wire is increased by 5%. So, new length is: L′=L+5100L=1.05L Let the new cross-sectional area be A′. On stretching the wire, its volume will remain constant. ⟹LA=L′A′ A′=A1.05 New resistance will become: R′=ρL′A′ R′=ρ(1.05LA/1.05) R′=1.052R=1.1025R Percentage increase in its resistance is, p=R′−RR×100 p=(1.1025−1)RR×100 p=10.25%
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