Picture 1 shows an inverted cone of height $h$ and radius $r$ . It contains water to a depth of $1/2 h$ . The volume of water it can hold is $480cm^3$ .
The cone is now inverted such that the liquid rests on the flat circular base of the cone. Find , in terms of $h$ , an expression for $d$ , the distance of the liquid surface from the Tip of the cone .
I'm completely stunned by this question.. I know it's got something to do with similarity . Can I get a hint on how to start ? Thanks in advance !
In the inverted cone, the section grows as the square of the height, so that, by integration, the volume grows as the cube, $V(h)=Vh^3/H^3$. Then in the straight cone, $V'(d)=V-Vd^3/H^3$.
By solving for $d$,
$$d=\sqrt[3]{H^3-h^3}.$$