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}{4 a}-\frac{21}{16}=0 $
$\frac{h}{a}=\frac{\frac{1}{4} \pm \sqrt{\frac{1}{16}+\frac{21}{4}}}{2} $
$\frac{h}{a}=\frac{1+\sqrt{85}}{8}$