A cube of wood, 20 cm on each side, floats in water so that 30% is above the sur

Question

A cube of wood, 20 cm on each side, floats in water so that 30% is above the surface of the water, and 70% is below. What is the density of the wood? (You must justify your answer) W hat mass of lead mass has to be placed on top of the block of wood, so that it will just be totally submerged? What volume of lead mass has to be placed on top of the block of wood, so that it will just be totally submerged?. What Force must be applied to the original cube of wood, so that it is 15% above the surface?

Explanation / Answer

wood of each side : 20 cm
volume = (0.2)^3 = 0.008 m^3

submerged 70 %

Hence volume submerged = 0.008*0.7 = 0.0056 m^3

a) Let density be d1
Hence

d1*0.008 = 0.0056*998
d1 = 698.6 kg/m³

b)
For completely wood getting submerged : 998*0.008 = 7.984 kg

Weight of wood = 698.6*0.008 = 5.5888

Mass of lead = 7.984 - 5.5888 = 2.3952

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