SSC JE Electrical Engineering 22 Jan 2018 Morning Paper

When wire is stretched to double its length, then the new resistance is given by
$\mathit{R}\prime ={\mathit{n}}^2\mathit{R}$
$\therefore \mathit{R}\prime ={\left(2\right)}^2\times 24$
$\therefore \mathit{R}\prime =\left(4\times 24\right)\mathit{\Omega }$
$\therefore \mathit{R}\prime =96\mathit{\Omega }$
Now when it is cut in two equal parts and connected in parallel then equivalent resistance
$\frac{1}{{\mathit{R}}_{\mathrm{eq}}}=\frac{1}{48}+\frac{1}{48}$ $=\frac{2}{48}$
$\therefore \frac{1}{{\mathit{R}}_{\mathrm{eq}}}=24\mathit{\Omega }$
Now $\mathit{I}=\frac{\mathit{V}}{\mathit{R}}=\frac{8}{24}=\frac{1}{3}\mathit{A}$ = 0.33 A