Let we have a wire of length
l, cross-sectional area A, and Resistivity
ρThen Resistance of wire is
R=ρ−(1)The volume of cylindrical wire is the product of its length and cross-sectional area.
V=I.AIf length is increased by 10 percent, the new length l' is
I′=∣+10% of I=1.1∣−(2)After stretching the cross-sectional area will also change, as volume remains constant. Let the new cross-sectional area be A'
V=I.A=I′A′... (3) Putting (2) in (3) we have
I⋅A=1.1I⋅A′A′=....(4)So, the Resistance of this stretched wire is
R′=ρ⋯ (5) Putting (2) and (4) in (5)
R′=ρ()⇒R′=ρ()⇒R′=ρ()⇒R′=1.21(ρ)−(6)Comparing
Eq(6) and
Eq(1) We get
R′=1.21RSo, the resistance increases
1.21 times.