Question icon
Grade 12Mechanics

1. Length of a conductor is increased by 1% .Calculate the percentage change in its resistance at constant volume

Profile image of jogesh
8 Years agoGrade 12
Answers icon

1 Answer

Profile image of Arun
8 Years ago
Given:
Let the initial length of wire=l1
Final length of wire after stretching=l1+(1/100)l1
=l1(1+1/100)
=( 101/100 )l1
= 101/100 l1
Area of initial wire=A1
Area of final wire = A2
Volume of wire remains constant.
l1A1=l2A2
A2=l1A1/l2
As we know that 
R = ρ l/A
R1/ R2 =( l1/A1 )/(l2/A2)
= l1A2/l2A1
= l1 (l1A1/l2)/l2A1
R1/R2 = l1^2/l2^2=101²/100²
R2=R1x101²/100²
(ΔR/R)×100=(R2-R1/R)×100=1.02%
The percentage increase in resistance IS 1.02%