A cube of 10cm and mass 0.5kg floats in a liquid with only one-fifth of it's height about the liquid surface.what is the Relative density?
1/5 of water?????
Just a guess though.
It certainly is a strange guess.
volume of cube = s^3
volume of water displaced = 4/5 s^3 = 0.8 s^3
Archimedes says weight of water displaced = buoyancy
if floating then buoyancy = weight
so
density water *.8 s^3 = density block * s^3
density block / density water = 0.8
ratio of
mass of cube = x s^3
mass of water displaced = 0.8
Yeah it was.
Typo. Meant to say 4/5 of the density of water, since only 1/5 isn't submerged.
To find the relative density of the cube, we need to consider the concept of buoyancy. The buoyant force acting on the cube is equal to the weight of the liquid displaced by the cube.
The weight of the cube is given by the formula:
Weight = Mass × Gravity
Given:
Mass of the cube = 0.5 kg
Gravity = 9.8 m/s^2 (approximate value on Earth)
Weight of the cube = 0.5 kg × 9.8 m/s^2 = 4.9 N
Since only one-fifth of the cube's height is above the liquid surface, the volume of the cube that is submerged in the liquid is (4/5) × height.
Given:
Side length of the cube = 10 cm = 0.1 m
Volume of the submerged cube = (4/5) × height × side length^2
To find the height, we need to subtract the height above the liquid surface from the total side length of the cube:
Height = side length - (1/5) × side length
Substituting the given values:
Height = 0.1 m - (1/5) × 0.1 m = 0.08 m
Volume of the submerged cube = (4/5) × 0.08 m × (0.1 m)^2 = 0.00064 m^3
Now, we need to find the density of the liquid in which the cube is floating. Let's assume the density of the liquid is ρ.
The weight of the liquid displaced by the cube is equal to the buoyant force acting on the cube:
Weight of liquid = Volume of displaced liquid × Density of liquid × Gravity
Given:
Weight of liquid = Weight of the cube = 4.9 N
Volume of displaced liquid = 0.00064 m^3
Gravity = 9.8 m/s^2
Plugging in the values:
4.9 N = 0.00064 m^3 × ρ × 9.8 m/s^2
Simplifying the equation:
ρ = 4.9 N / (0.00064 m^3 × 9.8 m/s^2)
Calculating the numerical value:
ρ ≈ 8039.06 kg/m^3
Therefore, the relative density of the cube is approximately 8039.06 kg/m^3.
volume of the cube= 10x10x10=1000cm³
mass of the cube= 0.5kg
density of d cube= 0.5/1000=0.0005kg/cm³
volume that floats in liquid=1/5*10*10*10=200cm³
mass of d cube in liquid= density*200cm³=0.0005kg/cm³*200cm³=0.1kg.
lost of mass=mass of the liquid=0.5kg-0.1kg=0.4kg.
R.D= mass of cube/mass of liquid=0.5/0.4=1.25