Question

By pulling a wire, its length increases by 0.1%, its resistance is the percentage change

Answer 1

Answer:

The resistance will increased by 0.2%

Explanation:

Let the initial length of the wire is L1, resistance is R1 and cross sectional area of the wire is A1.

And let assume that the new length of the wire after pulling is L2, reistance is R2 and the new cross sectional area of the wire is A2.

The length of the wire is increased by 0.1% (as the question).

So, L2 is equal to (0.1/100)L1+L1

So, L2 equal to 1.001L1

Let Initial volume of the wire is V1 and the new volume is V2

Since volume can't be changed, then

V1=V2

So, L1×A1=L2×A2

=> L1×A1= (1.001L1)×A2

=> A2= A1/1.001

So the resistance of the wire after pulling is-

R2 = ρL2/A2

= ρ*1.001L1/(A1/1.001)

= (1.001)²(ρL1/A1)

R2 = 1.002×R1

Therefore percentage change in resistance is={(New resistance-Initial resistance)/Initial resistance}*100%

=>(R2-R1)/R1×100%

=>(1.002R1-R1)/R1×100%

=>0.002×100%

=>0.2%