If an increase in the length of copper wire is 25% due to stretching, the percentage increase in its resistance will be:
The correct option is (B) 1%
Step 1, Given
Increase in length = 25%
Step 2,
We know,
As R = ρLA
25% increase in length means length is now L' = (L+25% of L)
i.e., L+25L100
L' = 1.25L
Decrease in area will be
L′×A′ = L.A (Since volume remains same)
1.25L×A′ = L.A
A′=11.25A
A' = 0.8A
Now, R=ρLA
putting L' = 1.25L and A' = 0.8A
R=ρ1.25L0.8A
So new resistance R′=1.56(ρLA)
or R' = 1.56R
% inrease in resistance (1.56R−R)R×100
= 56%
New resistance R' is 56% greater than old resistance R
Therefore for a 25% increase in length, the percentage increase in resistance is 56%.