SSC CGL Tier II 21 Sep 2014 Paper

(2) Let H and R be the height and radius of bigger cone respectively and h and r that of smaller cone. From triangles AOB and AMN.
∠A is common and MN || OB.
∴ Triangles AOB and AMN are similar,
$\therefore \frac{\mathit{A}\mathit{O}}{\mathit{A}\mathit{M}}=\frac{\mathit{B}\mathit{O}}{\mathit{M}\mathit{N}}$
$\Rightarrow \frac{30}{\mathit{h}}=\frac{\mathit{R}}{\mathit{r}}...\left(\mathit{i}\right)$
Volume of smaller cone $=\frac{1}{3}\mathrm{\pi }{\mathit{r}}^2\mathit{h}$
Volume of bigger cone $=\frac{1}{3}\mathrm{\pi }{\mathit{R}}^2\mathit{H}$
∴ According to the question,
$\frac{1}{3}\mathrm{\pi }{\mathit{r}}^2\mathit{h}=\left(\frac{1}{3}\mathrm{\pi }{\mathit{R}}^2\mathit{H}\right)\times \frac{1}{27}$
$\Rightarrow {\mathit{r}}^2\mathit{h}=\frac{{\mathit{R}}^2\mathit{H}}{27}$
$\Rightarrow 27{\mathit{r}}^2\mathit{h}={\mathit{R}}^2\mathit{H}$
$\Rightarrow \frac{27\mathit{h}}{\mathit{H}}=\frac{{\mathit{R}}^2}{{\mathit{r}}^2}$
$\Rightarrow \frac{27\mathit{h}}{\mathit{H}}={\left(\frac{30}{\mathit{h}}\right)}^2\left[\mathrm{From}\left(\mathit{i}\right)\right]$
$\Rightarrow \frac{27\mathit{h}}{\mathit{H}}=\frac{900}{{\mathit{h}}^2}$
$\Rightarrow 27{\mathit{h}}^3=900\mathit{H}=900\times 30$
$\Rightarrow {\mathit{h}}^3=\frac{900\times 30}{27}=1000$
$\Rightarrow \mathit{h}=\sqrt[3]{1000}=10\mathit{c}\mathit{m}$
∴ Required height = 30 – 10 = 20 cm