Question

A bottle in the shape of a right-circular cone with height h contains some water. When its base is placed on a flat surface, the height of the vertex from the water level is a units. When it is kept upside down, the height of the base from the water level is $\frac{a}{4}$ units. Then the ratio $\frac{h}{a}$ is

• A. $\frac{1+\sqrt{85}}{4}$
• B. $\frac{1+\sqrt{85}}{8}$
• C. $\frac{1+\sqrt{65}}{4}$
• D. $\frac{1+\sqrt{65}}{8}$

Answer 1

Answer: (B)

$\frac{r}{h}=\frac{x}{a}\Rightarrow x=\frac{ra}{h}$

$\frac{r}{h}=\frac{y}{h-\frac{a}{4}}$
$y=\frac{r}{h}\left(h-\frac{a}{4}\right)$
Equating volume of water in both cases
$\frac{1}{3}\left(\pi r^{2}h-\pi x^{2}a\right)=\frac{1}{3}\pi y^2\left(h-\frac{a}{4}\right)$
$\Rightarrow r^{2}h-r^{2}h-\frac{r^2{a}^2}{h}\cdot a$
$=\frac{r^2}{h^2}{\left(h-\frac{a}{4}\right)}^2\left(h-\frac{a}{4}\right)$
$\Rightarrow \frac{h^2}{a^2}-\frac{h}{4a}-\frac{21}{16}=0$
$\frac{h}{a}=\frac{\frac{1}{4}\pm \sqrt{\frac{1}{16}+\frac{21}{4}}}{2}$
$\frac{h}{a}=\frac{1+\sqrt{85}}{8}$