Question

Three sphere of radius 4cm each fit inside a tube calculate the percentage of the tube.

Answer 1

66.6% (see below)

Explanation:

Each sphere 4/3
`\pi`r²=4/3
`\pi`(4²)=268.08cm³

Assuming the tube is the same width and height as the three spheres,

V=
`\pi`r²h=
`\pi`(4²)24=1206.37

The three spheres are 268.08*3=804.24cm³

The tube is 1206.37cm³

I’m not sure if your question is asking the percentage of the tube that’s filled, or the percentage that’s unfilled.

Filled: 804.24÷1206.37=0.66666=66.6%

Unfilled: 1206.37-804.24=402.13

402.13÷106.37=0.33333=33.3%

Answer 2

Percentage of Tube Unfilled : 33 and 1/3%

Explanation:

It's most likely that we want to calculate the percentage of the tube unfilled, as the tube itself wouldn't be provided otherwise. We can start by calculating the volume of the tube --- (1)

The volume of a cylinder is represented by πr²h. Substituting we would receive π(4)²h. h is represented by 3 times the diameter of each sphere, as it is aligned such. Diameter = 2 * r = 2 * 4 = 8, so h = 3 * 8 = 24. Therefore Tube Volume = π(4)²(24) = π(16)(24) = 384π.

Now let's solve for the volume of a sphere, multiplying by three to receive the total volume of all 3 spheres --- (2)

Volume of 1 Sphere : 4 / 3πr³ = 4 / 3π(4)³ = 4 / 3π * 64 = 256 / 3π

Volume of all 3 Spheres : 256 / 3π * 3 = 256π

And now the volume of the tube that is unfilled, will be 384π - 256π = 128π. The total volume of the tube is 384π, so the percentage of the unfilled tube will be 128π / 384π or 128 / 384 = 0.333333333 * 100 = 33.33333...%. Therefore the percentage of the tube that is not filled will be 33.33% approximately, or 33 and 1/3% in exact terms.