Question

Let $V_1$ be the volume of a given right circular cone with $O$ as the centre of the base and $A$ as its apex. Let $V_2$ be the maximum volume of the right circular cone inscribed in the given cone whose apex is $O$ and whose base is parallel to the base of the given cone. Then, the ratio $V_2/V_1$ is

• A. $\frac{3}{25}$
• B. $\frac{4}{9}$
• C. $\frac{4}{27}$
• D. $\frac{8}{27}$

Answer 1

Answer: (C)

$V_1$ = Volume of cone $ABC$
$H$ = Height of cone $V_1$
$R$ = Radius of cone $V_1$
$V_2$ = Volume of cone inscribed in $V_1$
$(H-h)$ = Height of cone $V_2$
r = Radius of cone $V_2$

$\Delta A{O}^{\prime }Q$ and $\Delta AOC$ are similar
$\therefore \frac{A{O}^{\prime }}{AO}=\frac{O^{\prime }Q}{OC}$
$\Rightarrow \frac{h}{H}=\frac{r}{R}$
$\Rightarrow h=\frac{rH}{R}$
$\Rightarrow V_2=\frac{1}{3}\pi r^2(H-h)$
$\Rightarrow V_2=\frac{1}{3}\pi r^2\left(H-\frac{rH}{R}\right)$
$\Rightarrow \frac{d{V}_2}{dr}=\frac{1}{3}\pi H\left(2r-\frac{3r^2}{R}\right)$
For maxima or minima,
$\frac{d{V}_2}{dr}=0$
$\Rightarrow 2r-\frac{3r^2}{R}=0$
$\Rightarrow r=\frac{2R}{3}$
$\Rightarrow V_2=\frac{1}{3}\pi {\left(\frac{2}{3}R\right)}^2\left(H-\frac{2H}{3}\right)$
$=\frac{4\pi R^{2}H}{81}$
$V_1=\frac{1}{3}\pi R^{2}H$
$\therefore \frac{V_2}{V_1}=\frac{4\pi R^{2}H}{81}÷\frac{1}{3}\pi R^{2}H$
$=4:27=\frac{4}{27}$