Kindly solve problem 11:

Dear Student, 

(11.) Since the cube is completely immersed in the liquid, the density of the cube must be greater than or equal to the density of the liquid. The minimum condition is, the density of the cube is equal to the density of the liquid. We shall consider the minimum condition.

Let, density of the liquid = ρ_l

Density of water = 1 g/cc

Therefore, according to the given question relative density of the liquid is,

(ρ_l)/1 = 0.8

=> ρ_l = 0.8 g/cc = density of the cube

Volume of the cube is, V = x³= 125 cm³

=> x = 5 cm

Acceleration due to gravity, g = 980 cms^-2

Weight of the cube = (125)(0.8)(980) = 98000 Dyne

Let, the length of cube within water be 'd'

So, volume of water displaced by the cube 

= volume of the part of the cube within water

= x²d = 25d cm³

Therefore, buoyant force on the cube in water = (25d).(1).(9.8)= 24500d Dyne

Since the cube is floating,

Weight of the cube = Buoyant force

=> 98000 = 24500d

=> d = 4 cm

So, length of cube within water = 4 cm and the remaining 1 cm of the cube is above water.

Regards