Question

A 700 g solid cube of length 10 cm floats partially in water. The volume of the portion of the cube floating above the water level is

Answer 1

Hey there!

Given,
Mass of cube(m) = 700 g = 0.7 kg
Length of cube = 10 cm
And we know that density of water = 1000 kg/m³

We also know,
Density = mass / volume
So, volume = mass / density
                   = 0.7 kg / 1000 kg/m³ 
                   = 70000 m³
                   = 700 cm³
Now, 
Volume of cube = a³ = 10³ = 1000 cm³
Volume of cube's above portion = 1000 - 700 = 300 cm³

Hope It helps YoU!
                    

Answer 2

Hey! ! !

mate :-


Solution:

The weight of the cube is balanced by the buoyant force. The buoyant force is equal to the weight of the water displaced. If a volume V of the cube is inside the water, the weight of the displaced water = Vrg, Where r is the density of water,

Thus Vrg = (0.7kg) g.

Or, 

V = 0.7kg/p

= 0.7kg/1000kg/m*3

= 7 × 10 *-4 m*3

= 700 cm *3

The total volume of the cube = (10cm) 3 = 1000cm3.

The volume outside the water is 1000-700 = 300 cm3.