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Let $\mathrm{V}_{1}$ be the volume of a given right circular cone with $\mathrm{O}$ as the centre of the base and $\mathrm{A}$ as its apex. Let $\mathrm{V}_{2}$ be the maximum volume of the right circular cone inscribed in the given cone whose apex is $\mathrm{O}$ and whose base is parallel to the base of the given cone. Then the ratio $\mathrm{V}_{2} / \mathrm{V}_{1}$ is-
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Verified Answer
The correct answer is:
$\frac{4}{27}$
$\triangle \mathrm{ABC}$ and $\triangle \mathrm{AOP}$ are similar
$$
\begin{array}{c}
\frac{\mathrm{h}}{\mathrm{r}}=\frac{\mathrm{H}}{\mathrm{R}} \quad \Rightarrow \mathrm{h}=\frac{\mathrm{rH}}{\mathrm{R}} \\
\mathrm{V}_{2}=\frac{1}{3} \pi \mathrm{r}^{2}(\mathrm{H}-\mathrm{h})=\frac{\pi}{3} \mathrm{r}^{2} \mathrm{H}\left(1-\frac{\mathrm{r}}{\mathrm{R}}\right)=\frac{\pi \mathrm{H}}{3}\left(\mathrm{r}^{3}-\frac{\mathrm{r}^{3}}{\mathrm{R}}\right) \\
\frac{\mathrm{d} \mathrm{V}_{2}}{\mathrm{dr}}=2 \mathrm{r}-\frac{3 \mathrm{r}^{2}}{\mathrm{R}}=0 \quad \mathrm{r}=\frac{2 \mathrm{R}}{3} \\
\mathrm{~V}_{2 \max }=\frac{4 \pi \mathrm{R}^{2} \mathrm{H}}{81}
\end{array}
$$
$$
\begin{array}{l}
\mathrm{V}_{1}=\frac{\pi \mathrm{R}^{2} \mathrm{H}}{3} \\
\therefore \frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}=\frac{4}{27}
\end{array}
$$
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