The given question is not complete, the complete question is:
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.
Answer:
The correct answer is 0.29 cm.
Explanation:
To produce fake iron balls that is, formed of gold there is a need to make sure that the mass of the iron ball should be equivalent to the mass of the fake ball.
To calculate the volume of iron ball with the help of the given diameter, the formula to be used is 4/3πr³. The diameter of the spheres mentioned in the given question is 6 cm, so the radius will be 6/2= 3 cm.
Now, the volume of the iron ball would be = 4/3 × 3.14 × 3³ = 113.04 cm³
To determine the mass of the iron ball, the formula to be used is volume * density. Now putting the values we get,
113.04 × 5.15 g/cm³ = 582.156 grams (The density of the iron ore is 5.15 g/cm³)
Now, the mass of the gold ball should also be equal to 582.156 g. The density of the solid gold is 19.3 g/cm³, therefore, the volume of the gold ball by using the above formula would be,
mass of gold ball = Volume × density
Volume = mass of gold ball / density
Volume = 582.156 g / 19.3 g/cm³
= 30.1635 cm³
Thus, the volume of the hollow sphere would be 30.1635 cm³ with the outer radius as 3 cm, now there is a need to find the inner radius.
Volume of hollow ball = 4/3π [R³ -r³]
30.1635 cm^3 = 4/3 pie [3³ = r³]
30.1635 × 3/4 × 3.14 = 27-r³
7.2046 = 27-r³
r³ = 19.7954
r = 2.7051 cm
Hence, the thickness would be outer radius - inner radius = 3-2.7051 = 0.2949 cm or 0.29 cm.
