The complete question is One cubic meter of aluminum has a mass of 2.70 x 10^3 kg, and the same volume of iron has a mass of 7.86 × 10^3 kg. Find the radius of a solid aluminum sphere that will balance a solid iron sphere at radius 2 cm on an equal-arm balance. Solution with explanation please. TIA!

The volume of a sphere is directly proportional to the cube of its radius. So we can write V=kr³. This applies to both spheres, and k is constant, so if we have two volumes V₁ and V₂ and two corresponding radii, r₁ and r₂ we can write the equation V₁/V₂=r₁³/r₂³=(r₁/r₂)³. We don’t need the formula for the volume of a sphere.

The density of a sphere=mass/volume. So mass=volume×density. For iron mass is 7860V₁ kg and for aluminum it’s 2700V₂ kg, where we’ve used V₁ for the volume of the iron sphere and V₂ for the volume of the aluminum sphere. When their masses balance 7860V₁=2700V₂. So V₁/V₂=2700/7860=2.7/7.86=(r₁/r₂)³.

We know r₁=2cm so we have 2.7/7.86=(2/r₂)³. Now we take the cube root of both sides of the equation:

∛(2.7/7.86)=2/r₂ so r₂=2÷∛(2.7/7.86). We can use this formula to find r₂=2.856cm approx.

(We could have worked out the volume and mass of the iron sphere using the formula (4/3)πr³, and we could have converted cm to metres, but that would have involved more calculations and possible loss of accuracy and increased risk of error. Using proportions removes the need for superfluous calculations and reduces the risk of error.)

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