A cube of wood of density 0.78 g/cm³ is 11 cm on a side.When placed in water,what height of the block will float above the surface? (water density = 1.00 g/cm³) A) 8.6 cm B) 11 cm C) 2.4 cm D) 3.4 cm E) 1.4 cm

Question

Quizplus · Accepted Answer

To find the height of the cube that will float, we need to use the principle of buoyancy:

Buoyant force = weight of displaced water

The weight of the cube is:
Weight = Density x Volume x Gravity
Volume = Side^3 = 11^3 = 1331 cm^3
Weight = 0.78 g/cm^3 x 1331 cm^3 x 9.8 m/s^2 = 10,225.46 g

The weight of the displaced water is:
Weight = Density x Volume x Gravity
Volume = Area of base x Height = 11 cm x height
Weight = 1 g/cm^3 x (11 cm x height) x 9.8 m/s^2 = 107.8 x height g

Since the cube floats, the buoyant force is equal to the weight of the cube, so:

10,225.46 g = 107.8 x height g
height = 10,225.46 g / (107.8 g/cm^3 x 11 cm) = 2.4 cm

Therefore, the height of the cube that will float is 2.4 cm, which corresponds to choice C.

A Cube of Wood of Density 0