What is the ratio of empty to filled volume of the glass? The base diameter of a glass is $20$% smaller than the diameter at the rim. The glass is filled to half of the height. Then what is the ratio of empty to filled volume of the glass ?

Let $r$ the radius at the rim, then at the base the radius is $r_b=\frac{4}{5}r$. Supposing the glass has a truncated cone shape, then the radius at the middle of the glass is $r_m=\frac{9}{10}r$. The ratio you are looking for is
\begin{align}
\frac{V_{\text{empty}}}{V_{\text{filled}}}&=\frac{\frac{\pi h}{6}\left(r^2+rr_m+r_m^2\right)}{\frac{\pi h}{6}\left(r_m^2+r_mr_b+r_b^2\right)}\\[3pt]
&=\frac{r^2+\frac{9}{10}r^2+\frac{81}{100}r^2}{\frac{81}{100}r^2+\frac{36}{50}r^2+\frac{16}{25}r^2}\\[3pt]
&=\color{blue}{\frac{271}{217}}
\end{align}
Where $h$ is the height of the glass. The formula for the volume of a truncated cone is here
