Question

A wire 1 m long has a resistances of 1Ω. If it is uniformly stretched, so that its length increases by 25%, then its resistance will increase by:

• A. 25%
• B. 50%
• C. 56.25%
• D. 77.33%

Answer 1

The correct option is C

56.25%

We know that, resistance R = ρ$\frac{L}{A}$ and volume = L x A
where,
L = length of the wire before stretching and
A = area of the wire before stretching
L' = length of the wire after stretching and
A' = area of the wire after stretching
Given:
resistance of wire $R=1\Omega$
increase in length by percentage = 25%
new length, $L^{{}^{\prime }}$ = (L + 25% of L)
or $L^{{}^{\prime }}=1.25\times L$
Since the length of the wire increased and the total volume is constant
$L\times A$ = $L^{\prime }\times A^{\prime }$
$1\times A$ = $1.25\times A^{\prime }$
or $A^{\prime }=0.8A$
New resistance, $R^{{}^{\prime }}=\rho \frac{1.25L}{0.8A}$
or, $R^{{}^{\prime }}=1.5625R$
Increase in resistance = R' - R
Since $R=1\Omega$
Increase in resistance = 1.5625 - 1 = 0.5625
% increase in resistance = $\frac{R^{\prime }-R}{R}\times 100$
= $\frac{0.5625}{1}\times 100=56.25$
Therefore, increase in resistance = 56.25%.