# Answer to Question #21813 in Geometry for saketh

Question #21813

A closed vessel in the form of a right circular cone of height ‘h’ contains some water. When the vertex of the conical vessel is downward, water in it stands at 1/3rd of the height. It is now inverted so that the vertex is upward. At what height the water now stands ?

Expert's answer

When the vertex of the conical vessel isdownward:

V(water) = (1/3)* pi *(r/3)^2 * (h/3) = V0/27

V - volume

When the vertex of the conical vessel is upward:

V(water) = (1/3) pi * r^2*h - (1/3)*pi*(x*r)^2*(x*h)=V0(1-x^3)

x*h - heigth of cone without water

volume is constant:

1-x^3=1/27

x=(26/27)^(1/3) = 26^(1/3)/3

new h' = (1-x)*h

h'= (1-26^(1/3)/3)*h

V(water) = (1/3)* pi *(r/3)^2 * (h/3) = V0/27

V - volume

When the vertex of the conical vessel is upward:

V(water) = (1/3) pi * r^2*h - (1/3)*pi*(x*r)^2*(x*h)=V0(1-x^3)

x*h - heigth of cone without water

volume is constant:

1-x^3=1/27

x=(26/27)^(1/3) = 26^(1/3)/3

new h' = (1-x)*h

h'= (1-26^(1/3)/3)*h

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