Solving applied density problems 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 6 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 19.3 g/cm'). 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 g/cm'). 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. OP 0.0 X 5

thickness = 0.29 cm

Explanation:

In order to make fake iron ball [made of gold] we should get mass of fake ball

should be equal to that of Iron ball.So for that we should calculate

volume of iron ball using diameter given ;formula is 4/3 pi r^3

given d= 6 cm;so radius r= 6/2 = 3 cm

then volume of Iron ball = 4/3 *3.14* 3^3 = 113.04 cm^3

So mass of iron ball = volume x density = 113.04 * 5.15 g/cm^3 = 582.156 g

This must be the mass of gold ball ;now calculate volume of gold ball using its

density

volume of gold ball = mass of gold ball/density of gold ball = 582.156 g/19.3 g/cm^3

= 30.1635 cm^3

Now this must be the volume of hollow sphere whose outer radius R = 3cm

and inner radius r= ??

Volume of hollow ball = 4/3pi[R3-r^3]

30.1635 cm^3 = 4/3 pi [3^3-r^3]

30.1635* 3/4*3.14 = 27 - r^3

7.2046 = 27- r^3

r^3 = 19.7954

r= 2.7051 cm

So the thick ness = outer radius- inner radius = 3 - 2.7051 = 0.2949 cm

rounding to 2 significant figures

we get thickness = 0.29 cm

hey your not cool]