The length of a given cylindrical wire is increased to double of its original length. The percentage increase in the resistance of the wire will be ____________ %.

Solution

Physics sir · 4 steps

Stretching does not add metal: the wire gets longer and thinner together, so R gets hit twice.

1. Volume stays fixed. Stretching does not change the amount of metal, so A × l stays constant. If the length doubles to 2l, the cross-section area must halve to A/2.
2. Apply R = ρl/A. Resistance grows with length and falls with area. Here both changes push R up, and the resistivity ρ does not change because the material is the same.

   R' = (ρ (2l))/(A/2) = 4 (ρ l)/A = 4R

3. Percentage increase. The increase is 4R - R = 3R, so the percentage increase is 3R/R × 100 = 300%.
4. So. The resistance increases by 300%.

The common slip: 100. Keeping the cross-section area unchanged makes R only double, which gives 100, not 300. Stretching also thins the wire.

The idea to keep. When a wire is stretched by a factor n, the volume is fixed, so R ∝ l² and R becomes n² R. Percentage increase is new minus old, divided by old.

Answer from the NTA final answer key. Solution by PrepWiser.