Wire is stretched to increase in …
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Gaurav Seth 5 years, 2 months ago
Given:
Let the initial length of wire=l1
Final length of wire after stretching=l1+(10/100)l1
=l1(1+10/100)
=( 110/100 )l1
= 11/10 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=100/121
R2=R1x121/100
(ΔR/R)×100=(R2-R1/R)×100=21%
The percentage increase in resistance IS 21%
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