Answer:
1) Volume occupied by the spheres are equal therefore the three tanks contains the same volume of water
2) [tex]Amount \ of \, water \ remaining \ in \, the \ tank \ is \ \frac{x^3(6-\pi) }{6}[/tex]
Step-by-step explanation:
1) Here we have;
First tank A
Volume of tank = x³
The volume of the sphere = [tex]\frac{4}{3} \pi r^3[/tex]
However, the diameter of the sphere = x therefore;
r = x/2 and the volume of the sphere is thus;
volume of the sphere = [tex]\frac{4}{3} \pi \frac{x^3}{8}[/tex]= [tex]\frac{1}{6} \pi x^3[/tex]
For tank B
Volume of tank = x³
The volume of the spheres = [tex]8 \times \frac{4}{3} \pi r^3[/tex]
However, the diameter of the spheres 2·D = x therefore;
r = x/4 and the volume of the sphere is thus;
volume of the spheres = [tex]8 \times \frac{4}{3} \pi (\frac{x}{4})^3= \frac{x^3 \times \pi }{6}[/tex]
For tank C
Volume of tank = x³
The volume of the spheres = [tex]64 \times \frac{4}{3} \pi r^3[/tex]
However, the diameter of the spheres 4·D = x therefore;
r = x/8 and the volume of the sphere is thus;
volume of the spheres = [tex]64 \times \frac{4}{3} \pi (\frac{x}{8})^3= \frac{x^3 \times \pi }{6}[/tex]
Volume occupied by the spheres are equal therefore the three tanks contains the same volume of water
2) For the 4th tank, we have;
number of spheres on side of the tank, n is given thus;
n³ = 512
∴ n = ∛512 = 8
Hence we have;
Volume of tank = x³
The volume of the spheres = [tex]512 \times \frac{4}{3} \pi r^3[/tex]
However, the diameter of the spheres 8·D = x therefore;
r = x/16 and the volume of the sphere is thus;
volume of the spheres = [tex]512\times \frac{4}{3} \pi (\frac{x}{16})^3= \frac{x^3 \times \pi }{6}[/tex]
Amount of water remaining in the tank is given by the following expression;
Amount of water remaining in the tank = Volume of tank - volume of spheres
Amount of water remaining in the tank = [tex]x^3 - \frac{x^3 \times \pi }{6} = \frac{x^3(6-\pi) }{6}[/tex]
[tex]Amount \ of \ water \, remaining \, in \, the \ tank = \frac{x^3(6-\pi) }{6}[/tex].
