Question

A wire of resistance 1000 Ohm is uniformly
stretched so that its length increases by 5%
Assuming that the volume of the wire remains
the same on stretching, find the resistance of
the wire after stretching. Ans. 1102.50​

Answer 1

Given:-

→ Resistance of the wire = 1000 Ω

→ Percentage increase in length = 5%

→ Volume of the wire remains same.

To find:-

→ Resistance of the wire after stretching.

Solution:-

We know that :-

R = ρL/A.

Where :-

• R is resistance.

• ρ is resistivity.

• L is the length of the wire.

• A is the cross sectional area of the wire.

Case (1) :-

=> 1000 = ρL/A. -----(1)

______________________________

As the length of the wire increases 5% after stretching, thus new length (L') will be :-

=> L' = L + 5/100 L

=> L' = L + 0.05L

=> L' = 1.05L

Now as the length of the wire increases by 1.05 times, hence cross sectional area of the wire will decrease by 1.05 times.

∴ A' = A/1.05

Case (2) :-

=> R' = ρL'/A'

=> R' = ρ(1.05L)/(A/1.05)

=> R' = ρ(1.05L)(1.05)/A

=> R' = 1.1025ρL/A. -----(2)

On dividing eq.2 by eq.1, we get :-

=> R'/1000 = [1.1025ρL/A]/[ρL/A]

=> R'/1000 = 1.1025

=> R' = 1.1025(1000)

=> R' = 1102.50 Ω

Thus, resistance of the wire after stretching is 1102.50Ω .