6) A cube of wood having an edge dimension of 20.0 cm and a density of 650 kg/m3

Question

6) A cube of wood having an edge dimension of 20.0 cm and a density of 650 kg/m3 floats on water. (a) What is the distance from the horizontal top surface of the cube to the water level? (b) What mass of lead should be placed on the cube so that the top of the cube will be just level with the water surface? Ans: 7.0cm, 2.8 kg
6) A cube of wood having an edge dimension of 20.0 cm and a density of 650 kg/m3 floats on water. (a) What is the distance from the horizontal top surface of the cube to the water level? (b) What mass of lead should be placed on the cube so that the top of the cube will be just level with the water surface? Ans: 7.0cm, 2.8 kg
6) A cube of wood having an edge dimension of 20.0 cm and a density of 650 kg/m3 floats on water. (a) What is the distance from the horizontal top surface of the cube to the water level? (b) What mass of lead should be placed on the cube so that the top of the cube will be just level with the water surface? Ans: 7.0cm, 2.8 kg

Explanation / Answer

Let x cm = distance from top surface of the cube to the water level
=> 20 - x cm is inside water
Weight of cube = weight of water displaced
=> (20)^3 * 10^-6 * 650 * g = (20)^2 * (20-x) * 10^-6x1000 * g
=> 13000 = 1000 (20 - x)
=> 20 - x = 13
=> x = 7 cm.

2.
Lead wt. to be added
= Weight of water having same volume as of cube - weight of cube
= 20^3 * 10^-6 * 10^3 * g - 20^3 x 10^-6 * 650 * g
= (8 - 5.2) * 9.81 N
= 27.468 N

mg = 27.468 N

equivalent to m = 2.8 kg of lead

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