Help in calculating this tedious volume.

Question

The base diametre of glass is 20% smaller than the diametre of rim. The glass is filled to half the height.What is the ratio of empity to filled volume?

Answer 1

The volume of a truncated cone is given by: $$ V = \frac13 \pi h\ \big(R_{top}^2 + R_{top}\ R_{base} + R_{base}^2\big),$$ $h$ being the height.

You need to apply that to two different cones: one with the height $h$ of the glass and a second one with half the height $h'=h/2$. The smaller radius $R_{top}$ will be the same in both cases. You need to compute the bigger radius $R_{base}$ corresponding to height $h'$.

Answer 2

Hint let rim diameter $=d=x$ thus $r=x/2$ and thus base one =$d'=0.8(x)$ thus $r'=0.1(x)$ now volume of truncated cone or frustum is $$\frac{1}{3}\pi(r^2+r'^2+r.r')h$$ this is volume of empty one and filled one will be replacing $h$ by $h/2$ substitute $r,r'$ and get the ratio.