The total volume of the flask will be 50.06 [tex]\rm inches ^3[/tex] and if both the sphere and the cylinder are dilated by a scale factor of 2, the resulting volume would be '8' times the original volume.
Given :
Flask can be modeled as a combination of a sphere and a cylinder.
The volume of Sphere is given by the:
[tex]V_s = \dfrac{4}{3}\pi r^3[/tex]
Given - diameter of sphere = 4.5 inches. Therefore, radius is 2.25 inches.
Now, the volume of sphere of radius 2.25 inches will be:
[tex]V_s = \dfrac{4}{3}\times \pi\times (2.25)^3[/tex]
[tex]\rm V_s = 47.71\; inches^3[/tex]
The volume of Cylinder is given by the:
[tex]V_c = \pi r^2h[/tex]
Given - diameter of cylinder = 1 inches then radius is 0.5 inches and height is 3 inches.
Now, the volume of cylinder of radius 0.5 inches and height 3 inches will be:
[tex]V_c = \pi\times (0.5)^2 \times 3[/tex]
[tex]\rm V_c = 2.35\; inches^3[/tex]
Therefore the total volume of the flask will be = 47.71 + 2.35 = 50.06 [tex]\rm inches ^3[/tex].
Now, if both the sphere and the cylinder are dilated by a scale factor of 2 than:
Radius of sphere = [tex]2.25\times 2[/tex] = 4.5 inches
Radius of cylinder = [tex]0.5\times 2[/tex] = 1 inch
Height of cylinder = [tex]3\times 2[/tex] = 6 inches
Now, the volume of sphere when radius is 4.5 inches will be:
[tex]V_s' = \dfrac{4}{3}\times \pi \times (4.5)^3[/tex]
[tex]\rm V_s' = 381.70\; inches ^3[/tex]
And the volume of cylinder when radius is 1 inch and height is 6 inches will be:
[tex]V_c' = \pi \times (1)^2\times 6[/tex]
[tex]\rm V_c'=18.85\;inches^3[/tex]
Therefore the total volume of the flask after dilation by a scale factor of 2 will be = 381.70 + 18.85 = 400.55 [tex]\rm inches ^3[/tex].
Now, divide volume with dilation by theorginal volume of the flask.
[tex]\dfrac{400.55}{50.06}=8[/tex]
Therefore, if both the sphere and the cylinder are dilated by a scale factor of 2, the resulting volume would be '8' times the original volume.
For more information, refer the link given below:
https://brainly.com/question/15861918
