Mensuration Practice Test 9

Volume of the hemispherical cup = $\frac{2}{3}\mathit{\pi }{\mathit{r}}^2$
= $\frac{2}{3}\times \frac{22}{7}$ × 4 × 4 × 4 = 134.09 $\mathit{c}{\mathit{m}}^3$
Volume of the cone = $\frac{1}{3}\mathit{\pi }{\mathit{r}}^2\mathit{h}$
= $\frac{1}{3}\times \frac{22}{7}$ × 8 × 8 × 16 = 1072.5
Volume of the empty cone = 1,072.75 − 134.09
= 938.67 $\mathit{c}{\mathit{m}}^3$
Percentage of the empty cone = $\frac{93.67}{1072.75}\times 100$
= 87.5 %