Question #265980

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.


1
Expert's answer
2021-11-15T16:36:50-0500

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

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

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


xh=RH\dfrac{x}{h}=\dfrac{R}{H}

The volume of the water is


Vwater=13πx2hV_{water}=\dfrac{1}{3}\pi x^2h

Given h=1mh=1m


Vwater=13π(RH)2(1m)3V_{water}=\dfrac{1}{3}\pi (\dfrac{R}{H})^2(1m)^3

When the vessel was turned over


Vwater=13πR2H13π(RH)2(133m)3V_{water}=\dfrac{1}{3}\pi R^2H-\dfrac{1}{3}\pi (\dfrac{R}{H})^2(\sqrt[3]{13}m)^3

Then


Vwater=13π(RH)2(1m)3V_{water}=\dfrac{1}{3}\pi (\dfrac{R}{H})^2(1m)^3

=13πR2H13π(RH)2(133m)3=\dfrac{1}{3}\pi R^2H-\dfrac{1}{3}\pi (\dfrac{R}{H})^2(\sqrt[3]{13}m)^3

1=H3131=H^3-13

H=143 mH=\sqrt[3]{14}\ m

H=2.41 mH=2.41\ m

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