Xavier Aptitude Test 2012 Paper

(e) $\frac{4}{3}\mathrm{\pi }{\mathit{R}}^3=1000.\frac{4}{3}\mathrm{\pi }{\mathit{r}}^3$
$\mathit{R}=10\mathit{r}$
$\therefore \mathrm{}\mathit{r}=$ radius of smaller sphere $=1\mathrm{}\mathit{c}\mathit{m}$
$\mathit{R}=$ radian of spherical metal
Initial Surface area of metal $=4\mathit{\pi }{\mathit{R}}^2=4\mathrm{\pi }\times 100=400\mathit{\pi }$
Final surface area of 1000 smaller sphere $=1000.4\mathrm{p}\mathit{\pi }{\mathit{r}}^2$
$=1000\times 4\mathrm{\pi }=4000\mathrm{\pi }$
$\therefore$ Increase in surface area $=4000\mathrm{\pi }-400\mathrm{\pi }=3600\mathrm{\pi }$
$\mathrm{}\frac{3600\mathrm{\pi }}{400\mathrm{\pi }}=9$ times
Hence correct option is (e)