Mensuration Practice Test 6

Let the radius of the smaller hemisphere = r
Radius of the larger hemisphere = R
$\frac{\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{smaller}\mathrm{hemisphere}}{\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{larger}\mathrm{hemisphere}}$ = $\frac{\frac{2}{3}\mathit{\pi }{\mathit{r}}^3}{\frac{2}{3}\mathit{\pi }{\mathit{R}}^3}$
= $\frac{6.4\times 1000}{21.6\times 1000}$ = $\frac{{\mathit{r}}^3}{{\mathit{R}}^3}$
= $\frac{4}{6}$ = $\frac{\mathit{r}}{\mathit{R}}$ = $\frac{2}{3}$ = $\frac{\mathit{r}}{\mathit{R}}$
Inner curved surface area
= $\frac{\mathrm{smaller}\mathrm{hemisphere}}{\mathrm{inner}\mathrm{curved}\mathrm{surface}\mathrm{area}}$ = $\frac{2\mathit{\pi }{\mathit{r}}^2}{2\mathit{\pi }{\mathit{R}}^2}$ of larger hemisphere
= $\frac{{\left(2\right)}^3}{{\left(3\right)}^2}$ = $\frac{4}{9}$
So, required ratio = 4 : 9