Question

If an increase in the length of copper wire is 25% due to stretching, the percentage increase in its resistance will be:

• A. 1%
• B. 0.2%
• C. 1%
• D. 2%

Answer 1

The correct option is A

1%

As R = ρ$\frac{L}{A}$
25% increase in length means length is now L' = (L+25% of L)
i.e., $L+\frac{25L}{100}$
L' = 1.25L
decrease in area will be
$L^{\prime }\times A^{\prime }$ = L.A (Since volume remains same)
$1.25L\times A^{\prime }$ = L.A
$A^{\prime }=\frac{1}{1.25}$A
A' = 0.8A

Now, $R=\rho \frac{L}{A}$
putting L' = 1.25L and A' = 0.8A
$R=\rho \frac{1.25L}{0.8A}$
So new resistance $R^{\prime }=1.56\left(\rho \frac{L}{A}\right)$
or R' = 1.56R
% inrease in resistance $\frac{(1.56R-R)}{R}\times 100$
= 56%
New resistance R' is 56% greater than old resistance R
Therefore for 25% increase in length, percentage increase in resistance is 56%.