SSC CGL Tier II 12 Jan 2017 Paper 1

(3) Volume of copper sphere $=\frac{4}{3}\mathrm{\pi }{\mathit{r}}^3$ $=\frac{4}{3}\mathrm{\pi }{\left(21\right)}^3\mathrm{cu}.\mathrm{cm}$
Volume of cylindrical rod
$=\mathrm{\pi }{\mathit{R}}^2\mathit{H}=\mathrm{\pi }{\mathit{R}}^2\times 7\mathrm{cu}.\mathrm{cm}.$
$\mathrm{\pi }{\mathit{R}}^2\times 7=\mathrm{}\frac{4}{3}\mathrm{\pi }\times 21\times 21\times 21$
$\Rightarrow {\mathit{R}}^2=\frac{4}{3}\times \frac{21\times 21\times 21}{7}$
$\therefore \mathit{R}=\sqrt{4\times 21\times 21}=2\times 21$ = 42 cm.
Surface area of sphere $=4\mathrm{\pi }{\mathit{r}}^2$ $=4\mathrm{\pi }{\left(21\right)}^2$
Total surface area of the rod $=\mathrm{\pi }\mathit{R}\left(\mathit{R}+\mathit{H}\right)$
= 2π × 42 (42 + 7)
= 2π × 42 × 49 sq. cm.
∴ Required ratio $=\frac{2\mathrm{\pi }\times 42\times 49}{4\mathrm{\pi }\times 21\times 21}=7:3$