Mensuration Practice Test 4

Radius of the cone = 2 cm
Height = 6 cm.
Radius of the cone R = 4 cm
Height remain fixed = 6 cm.
Volume of the cone = $\frac{1}{3}\mathit{\pi }{\mathit{r}}^2\mathit{h}$ = $\frac{\mathit{\pi }{\left(2\right)}^{26}}{3}$ = 8π
Volume of the new cone = $\frac{1}{3}\mathit{\pi }{\mathit{R}}^2\mathit{h}$
= $\frac{1}{3}\mathit{\pi }{\left(4\right)}^{2}6$
= 32π
Volume of the new cones is four times of the previous volume.