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 4cm 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 19.3/gcm3). He suggests forming the gold into hollow balls instead (see sketch at right), so that the fake "iron ore" has the same density as real iron ore (5.15/gcm3).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.

To solve this problem, we need to find the thickness of the gold shell that will give the sphere the same density as iron ore.

First, let's calculate the volume of the sphere using the formula for the volume of a sphere, which is V = 4/3 * π * r³. The radius (r) is half the diameter, so r = 2 cm.

V = 4/3 * π * (2 cm)³ = 33.51 cm³

The mass of the iron ore that the sphere should have is the volume times the density of iron ore:

Mass = V * density_iron = 33.51 cm³ * 5.15 g/cm³ = 172.58 g

Now, let's denote the thickness of the gold shell as t. The outer radius of the gold shell is 2 cm, and the inner radius is (2-t) cm. The volume of the gold shell is the difference between the volumes of the outer and inner spheres:

V_gold = 4/3 * π * (2 cm)³ - 4/3 * π * ((2-t) cm)³

The mass of the gold shell is the volume times the density of gold:

Mass_gold = V_gold * density_gold

We want the mass of the gold shell to be equal to the mass of the iron ore, so we set them equal to each other and solve for t:

172.58 g = (4/3 * π * (2 cm)³ - 4/3 * π * ((2-t) cm)³) * 19.3 g/cm³

Solving this equation for t, we get:

t = 2 cm - ((172.58 g / (4/3 * π * 19.3 g/cm³))^(1/3))

Calculating this gives:

t = 2 cm - 1.24 cm = 0.76 cm

So, the required thickness of the walls of each hollow lump of "iron ore" is approximately 0.76 cm.