A wire of length 20 m and resistance 1 omega is stretched uniformly so that its length is increased by 25%. Find the increase in resistance.

$Details And Assumptions$

The answer is not 100

Enter your answer is the number without the percentage sign. E.g. If the answer is 50% then enter your answer as 50

The answer is 56.25.

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Resistance $R$ of a conductor wire is given by:

$R = \dfrac {\rho l}{A}$

where

For a wire with a constant volume $V = Al$ . This implies that:

$R = \dfrac {\rho l}{A} = \dfrac {\rho l^2}{V}\quad \Rightarrow R \propto l^2$

Therefore, for an increase of $25\%$ in $l$ , the corresponding increase in $R$ is:

$\dfrac {R_1}{R_0} = \dfrac { \left( l_0 +\frac {1}{4} l_0 \right)^2} {l_0^2} = \left ( \dfrac {5}{4} \right)^2 = \dfrac {25}{16}$

The percentage increase in resistance is:

$\Delta R = \left( \dfrac {25}{16} - 1 \right) \times 100 = \dfrac {900}{16} = \boxed {56.25} \%$