Volume of water in a cone Let, slant height of a cone be 6cm, and radius be 3cm and the cone be uniform. Let a uniform & solid sphere of radius 1cm be put in the cone & fill the cone by water. What is the minimum volume of water such that the sphere is under water?
I find that the height of the cone is $\sqrt(27)$cm & I can not proceed further.

You can easily compute the demi-angle $\alpha$ of the cone.
$\cos \alpha=\dfrac {\sqrt{27}}{6}$
Consider that the sphere is in contact with the edge of the cone. Thus the distance between the center of the sphere $O$ with the edge is $1$, radius of the sphere. 
Let's call $d$ the distance between the apex of the cone and the center of the sphere.
Then $\sin \alpha=\dfrac 1d$
From there you deduce $d$. In order to cover the sphere, you need to be at the upper edge, whose height id $d+1$.
You just need to compute the volume of the cone of demi-angle $\alpha$, of height $d+1$, and then substract the volume of the sphere. 
EDIT: based on a comment, I used the height being $5$, which is wrong... height is $\sqrt{27}$
