Question

One cubic meter (1.00 m³) of aluminum has a mass of 2.70  10³ kg, and the same volume of iron has a mass of 7.86  10³ kg. Find the radius of a solid aluminum sphere that will balance a solid iron sphere of radius 2.02 cm on an equal-arm balance.

Answer 1

On an equal-arm balance:

Wa = Wi

Ma*g = Mi*g

Ma = Mi

here, Ma = mass of aluminium = a*Va

Mi = mass of iron = i*Vi

given, i = density of iron = 7.86*10^3 kg/m^3

a = density of aluminium = 2.70*10^3 kg/m^3

Vi = volume of iron = (4/3)*pi*Ri^3

Va = volume of aluminium = (4/3)*pi*Ra^3

Ri = radius of iron sphere = 2.02 cm

So,

a*(4/3)*pi*Ra^3 = i*(4/3)*pi*Ri^3

Ra^3 = (i/a)*Ri^3

Ra^3 = [(7.86*10^3)/(2.70*10^3)]*2.02^3

Ra = 24^(1/3)

Ra = 2.88 cm

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