Question

In order to increase the resistance of a given wire of uniform cross section to four times its value, a fraction of its length is stretched uniformly till the full length of the wire becomes $\frac{3}{2}$ times the original length. If the fraction of wire that is streched is found to be $\frac{n}{16}$. Then the value of $n$ is ?

Answer 1

$Ax=A^{\prime }(0.5\ell +x)$

$A^{\prime }=\frac{Ax}{0.5\ell +x}$

$\Rightarrow \frac{4p\ell }{A}=\frac{\rho (\ell -x)}{A}+\frac{\rho (0.5\ell +x)}{A^{\prime }}$

$\Rightarrow \frac{4\rho \ell }{A}=\frac{\rho (\ell -x)}{A}+\frac{\rho (0.5\ell +x{)}^2}{Ax}$

$\Rightarrow 4\ell x=\ell x-x^2+(0.5\ell {)}^2+\ell x+x^2$

After solving $x=\left(1/8\right)\ell$
Thus, $\frac{n}{16}$ = $\frac{1}{8}$
$⟹$ $n$ = 2