Question

An electric wire is stretched to increase its length by 25%.by what %will the resistance be increased and what will be increase in its resistivity?

Answer 1

R = ρL/A
ρ is resistivity of the material
L is length in meters
A is cross-sectional area in m²

we know the resistance will go up, and I think it will go up by about 50%, but need the math.

first find out how much the cross-sectional area changes.
volume of a cylinder = cross-sectional area x h
V = Ah
initial volume V1 = A1h1
final volume V2 = A2h2
volume does not change
A1h1 = A2h2
A1/A2 = h2/h1 = 1.25
so cross-sectional area will go down by 1.25

now resistance calculation
initial resistance R1 = ρL1/A1
final resistance R2 = ρL2/A2
R2/R1 = ρL2/A2 / ρL1/A1
R2/R1 = (ρL2/A2) * (A1/ρL1)
R2/R1 = (L2/L1) * (A1/A2)
so the ratio is the product of the two
R2/R1 = 1.25 * 1.25
R2/R1 = 1.56

resistance goes up by 56%

Answer 2

Answer:

Explanation:

Explanation:

Formula used, R= PL/A

As length is increased by stretching the wire , therefore the volume of the wire won't change. Therefore, final volume=Initial volume

Hence, AL=A'L'

Final length= {1+(25/100)} Initial length

Hence, final length L'={125/100}L...(1)

R=pL/A

Multiply den and num by L

Hence, R= pL^2/V. (AL=V)

Finally, R'= pL'^2/V

Put the value of L' from eq1

R'= 1.5625R

Or resistance increase by

(156.25)R-R/(R)×100%= 56.25%

There will be no change in resistivity as resistivity is independent of the dimensions of wire.