Answer to Question #265980 in Math for S.S

Question

Water is poured into a vessel of the form of a straight circular cone. If the vessel is installed with the <<sharp>> end down then the distance from the water level to the base of the cone is 1 m. When the vessel was turned over, it turned out that the distance from the water level to the <<sharp>> end of the vessel is root(3, 13) * m . Find the height of the vessel; give your answer in meters, rounded to two decimals if needed. The volume of a cone can be found by the V = 1/3 * S * h formula , where S is the area of its base and h is its height.

Answer 1

Let $R=$ the base radius of the cone, $H=$ the height of the cone.

If the vessel is installed with the "sharp" end down

$x=$ the base radius of the "water"cone, $h=$ the height of the "water"cone.

$$\dfrac{x}{h}=\dfrac{R}{H}$$

The volume of the water is

$$V_{water}=\dfrac{1}{3}\pi x^2h$$

Given $h=1m$

$$V_{water}=\dfrac{1}{3}\pi (\dfrac{R}{H})^2(1m)^3$$

When the vessel was turned over

$$V_{water}=\dfrac{1}{3}\pi R^2H-\dfrac{1}{3}\pi (\dfrac{R}{H})^2(\sqrt[3]{13}m)^3$$

Then

$$V_{water}=\dfrac{1}{3}\pi (\dfrac{R}{H})^2(1m)^3$$

$$=\dfrac{1}{3}\pi R^2H-\dfrac{1}{3}\pi (\dfrac{R}{H})^2(\sqrt[3]{13}m)^3$$

$$1=H^3-13$$

$$H=\sqrt[3]{14}\ m$$

$$H=2.41\ m$$