Question

A wire is stretched $\frac{6}{5}$ times of its original length. Calculate its new resistance, if initial resistance was $25$ $\Omega$.

Answer 1

$R_1=\frac{{\rho }_1{l}_1}{A_1}$

$l_2=\frac{6}{5}{l}_1$

$R_1=25$

$R_2=\frac{{\rho }_2{l}_2}{A_2}$

As, ${\rho }_2={\rho }_1$

And $A_2=A_1$

$\frac{R_2}{R_1}=\frac{l_2}{l_1}$

$⟹R_2=\frac{6}{5}\times 25$

$⟹R_2=30$