RRB NTPC Mains or Stage 2 Model Paper 5 with solutions

(d) Volume of wire $=\mathit{\pi }{\mathit{r}}^2\mathit{h}$
New radius of the wire $=\frac{\mathit{r}\times 90}{100}=\frac{9\mathit{r}}{10}$
Let new length of the wire be L.
∴ Volume of new wire
$=\mathit{\pi }{\left[\frac{9\mathit{r}}{10}\right]}^2\times \mathit{L}=\frac{81}{100}\mathit{\pi }{\mathit{r}}^2\mathit{L}$
According to question,
$\mathit{\pi }{\mathit{r}}^2\mathit{h}=\frac{81}{100}\mathit{\pi }{\mathit{r}}^2\mathit{L}\Rightarrow \mathit{L}=\frac{100}{81}\mathit{h}$
Increase in length =$\frac{100}{81}\mathit{h}-\mathit{h}=\frac{19}{81}\mathit{h}$
Percent increase =$\frac{\frac{19}{81}\mathit{h}}{\mathit{h}}\times 100\%=23.46\%=$
= 23% (approx)