Two paper drink cups are shaped like cones. The small cone can hold 6 oz of water. The large cone is 4/3 the height and 4/3 the diameter of the small cone. Which of these could be the amount of water the large cone holds?

Amount of water contained in the cone shaped large cup will be given by Option (B). 14 oz
V = [tex]\frac{1}{3}\pi(r)^2h[/tex]
Here, r = radius of the circular base of the cone
h = Height of the cone
Let the height of the small cone = h cm
And the radius of the small cone = r cm
Therefore, volume of the small cone = [tex]\frac{1}{3}\pi(r)^2h[/tex]
Radius of the large cone = [tex](\frac{4}{3}r)[/tex] = [tex]\frac{4}{3}r[/tex]
Height of the large cone = [tex]\frac{4}{3}h[/tex]
Volume of the large cone = [tex]\frac{1}{3}(\pi)(\frac{4}{3}r )^2(\frac{4}{3} h)[/tex]
= [tex]\frac{1}{3}(\pi )(\frac{16}{9}r^2 )(\frac{4}{3}h )[/tex]
= [tex]\frac{1}{3}(\frac{64}{27})(\pi r^2h)[/tex]
= [tex](\frac{64}{27})(\frac{1}{3} \pi r^2h)[/tex]
= (64/27)×(Volume of the small cone)
Since, volume of the small cone = amount of water in the cuo
= 6 oz
Therefore, amount of water contained in large cone = [tex]\frac{64}{27}\times 6[/tex]
= 14.22 oz
≈ 14 oz
Hence water in the large cone will be Option (2). 14 oz
Learn more about the volume of a cone here,
https://brainly.com/question/14933149?referrer=searchResults