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A rectangular bar of soap (density 800 kg/m³) floats at the interface between water (density 1000 kg/m³) and oil (density 300 kg/m³) added above. Oil is added slowly until the top surface of the oil is level with the top surface of the soap. If x is the thickness of the oil layer and L is the total thickness of the soap bar, find x/L.
- 2/10
- 2/7
- 3/10
- 3/8
Correct answer: 2/7
Solution
Weight of soap per unit cross-section = 800*L*g. Buoyancy = 1000*d*g + 300*(L-d)*g. Setting equal: 800L = 1000d + 300(L-d) = 700d + 300L, so 500L = 700d, d = 5L/7. Then x = L - d = 2L/7, giving x/L = 2/7.
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