20 Followers Problem!!

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

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

where

• $\rho \space (\Omega m^{-1})$ is the resistivity of the material of the wire
• $l\space (m)$ is the length of the wire
• $A\space (m^{-2})$ is the cross-sectional area of the wire.

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:

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

When a wire is stretched n times resistance new resistance= n^2 x initial resistance.

After stretching , resistance of the wire is R=1.25x1.25x1=1.5625

Therefore increase %= 0.5625/1 x 100 = 56.25%