What is the height of the conical bottle?

Consider the image shown above.

Let the height of the cone be "h" and "R" be the radius of the cone. When the bottle is rested on its flat base, the volume of water is given by

Volume of the cone - volume of empty space above the water level

\(\frac{\pi}{3} R^2 h\) - \(\frac{\pi}{3} r_1^2 (8)\)

Now by similar triangle concept:

\(\frac{r_1}{8} = \frac{R}{h}\)

\(r_1 = \frac{8R}{h}\)

Sub \(r_1\) value in the above equation, we get

volume of water = \(\frac{\pi}{3} R^2 h - \frac{\pi}{3} (\frac{8R}{h})^2 (8)\)

\(\frac{\pi}{3} R^2 (h - \frac{512}{h^2})\) ......................................... (1)

Now consider the cone is tilted upside down. Volume of water is

\(\frac{\pi}{3} (r_2)^2 (h-2)\)

Again by similar triangle concept:

\(\frac{r_2}{h-2} = \frac{R}{h}\)

\(r_2 = \frac{(h-2) R}{h}\)

Substituting \(r_2\) in the above equation of volume of water when cone is tilted,

volume of water = \(\frac{\pi}{3} (\frac{(h-2) R}{h})^2 (h-2)\)

\(\frac{\pi}{3} R^2 \frac{(h-2)^3}{h^2}\) ..................................... (2)

Since volume of water in both the cases is equal, (1) = (2)

\(\frac{\pi}{3} R^2 (h - \frac{512}{h^2})\) = \(\frac{\pi}{3} R^2 \frac{(h-2)^3}{h^2}\)

\(h - \frac{512}{h^2} = \frac{(h-2)^3}{h^2}\)

\(h^3 - 512 = (h-2)^3\)

\(h^3 - 512 = h^3 - 6h^2 + 12h - 8\)

\(6h^2 - 12h - 504 = 0\)

\(h^2 - 2h - 84 = 0\)

Since \(h > 0 , h \approx {10 cm}\)

OPTION : C