Question

Mr. Auric Goldfinger, criminal mastermind, intends to smuggle several tons of gold across international borders by disguising it as lumps of iron ore. He commands his engineer minions to form the gold into little spheres with a diameter of exactly
8 cm and paint them black.

However, his chief engineer points out that customs officials will surely notice the unusual weight of the "iron ore" if the balls are made of solid gold (density
). She suggests forming the gold into hollow balls instead , so that the fake "iron ore" has the same density as real iron ore
One of the balls of fake "iron ore," sliced in half.

Calculate the required thickness of the walls of each hollow lump of "iron ore." Be sure your answer has a unit symbol, if necessary, and round it to 2
significant digits.

Answer 1

The thickness of the walls of each hollow lump of iron ore is: 0.295cm. This is determined using the knowledge of the Volume of a Hollow Sphere.

What is the Volume of a Hollow Sphere?

The formula for the Volume of a Hollow Sphere is give as:

Volume of a Hollow = 4/3 R³ – 4/3 r³

Step 1 - Find the apparent volume of 8cm diameter sphere including the wall the the hollow inside

V = 4/3 r³

V = (4/3) (8)³

V = (4/3) * (22/7) * 512

V = 2145.52cm³

Step 2 - Determine the mass of Gold required for the sphere to achieve the density of iron ore if the apparent volume is as given in Step 1 above.

Recall that Mass = Density * Volume

Mass = 5.15 *  2145.52

Mass = 11049.45

Step 3 - Find volume of gold based on mass calculated in Step 2 above

Recall that :

Volume = Mass/Density

Thus,

V = 11049.4/19.3

Thus, Volume = 572.51cm³

Step 4 - Find the width of the wall of a hollow sphere for the volume realized in step 3

Recall the initial formula:

Volume of a Hollow = 4/3 R³ – 4/3 r³

Thus, 572.51 = (4/3) (8³ - (8 - t)³)

572.51 = (4/3) *(22.7) *(8³ - (8 - t)³)

572.51 = (88/21) *(8³ - (8 - t)³)

Multiply both sides by 21 and we have

88 (512 - (8 - t)³) = 12,022.71

Simplify further
(8 - t)³ = 357.38

t = 0.9035cm

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