One cubic meter (1.00 m^3) of aluminum has a mass of 2.70 multiplied by 10^3 kg, and the same volume of iron has a mass of 7.86 multiplied by 10^3 kg. Find the radius of a solid aluminum sphere that will balance a solid iron sphere of radius 1.40 cm on an equal-arm balance.

Answer 1

To have the same weight and mass, the product of Radius^3 and density must be the same for the aluminum and iron spheres.

Let subscript 1 be Al and subscript 2 be iron.

R1^3*(density1) = R2^3*(density2)

(R1/R2)^3 = 7.86/2.70 = 2.911
R1/R2 = cube root of 2.911 = 1.428
R1 = 1.40*1.428 = 2.00 meters
(using three significnant figures)