Question

A cube of wood with dimension 20 cm and density 650
kg/m^{3} floats in water.

a.)What is the distance from the top surface of the cube to the surface of the water?

b.)How much mass of lead is required to be put on cube so that the top of the cube is level with the surface of the water?

Answer #1

Volume of wood cube = 0.20^3 = 0.008 m^3

Density = 650 Kg/m^3

Mass = Volume*density = 0.008*650 = 5.2 Kg

a) For the wood cube to float in water, weight of the cube is equal to the buoyant force acting on it. But Buoyant force is equal to weight of water displaced. Therefore, weight of cube = weight of water displaced. Let h be the height of the top surface of the cube to the surface of the water. Therefore, (0.2 - h) is submerged in the water.

mg = pwater.g.V

=> m = pwater*V

5.2 = 1000*0.20^2 *(0.20 - h)

2.8 = 40h

=> h = 0.07 m = 7 cm

b) To be just level, the weight of lead + wood block should be equal to the weight of the displaced water.

=> mass of wood +lead block = mass of displaced water

but since volume is same in both cases. Therefore

density of wood + lead block = density of water

mass = volume*density = 0.008*1000 = 8 Kg total mass

Hence mass of lead = 8 - 5.2 = 2.8 Kg

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