SSC CGL Tier II 25 Oct 2015 Paper 1

(4) Volume of sphere $=\frac{4}{3}\mathit{\pi }{\mathit{r}}^3$
∴ Total volume of both spheres $=\frac{4}{3}\mathrm{\pi }\left({\mathit{r}}_1^3+{\mathit{r}}_2^3\right)$
$=\frac{4}{3}\mathrm{\pi }\left(1^3+6^3\right)$
$=\frac{4}{3}\mathrm{\pi }\left(1+216\right)$
$=\left(\frac{4\mathrm{\pi }}{3}\times 217\right)\mathrm{cu}.\mathrm{cm}$
If the internal radius of hollow sphere = r cm, then
∴ Volume of the iron of this sphere $=\frac{4}{3}\mathrm{\pi }\left(9^3-{\mathit{r}}^3\right)\mathrm{cu}.\mathrm{cm}$
According to the question,
$\frac{4}{3}\mathrm{\pi }\left(9^3-{\mathit{r}}^3\right)=\frac{4\mathrm{\pi }}{3}\times 217$
$\Rightarrow 729-{\mathit{r}}^3=217$
$\Rightarrow {\mathit{r}}^3=729-217=512$
$\Rightarrow {\mathit{r}}^3={\left(8\right)}^3$
$\Rightarrow \mathit{r}=8\mathit{c}\mathit{m}$
∴ Required thickness = 9 – r = 9 – 8 = 1 cm.