An ice cube floats with 1/9 of its volume below the surface of water, calculate the density of ice.
Vb = Di/Dw * Vi = Vi/9.
Di/1000 * Vi = Vi/9,
Divide both sides by Vi:
Di/1000 = 1/9, Di = 1000/9 = 111.11 kg/m^3 = Density of the ice cube.
To calculate the density of ice, we need to use the concept of buoyancy.
When an object floats in a fluid, the buoyant force acting on the object is equal to the weight of the fluid it displaces. In this case, the buoyant force acting on the ice cube is equal to the weight of the water it displaces. Since the ice cube floats with 1/9 of its volume below the surface, we can assume that 1/9 of the ice cube's volume is submerged in the water, and 8/9 of its volume is above the surface.
Let's denote the density of the ice cube as ρ_ice, and the density of water as ρ_water. The buoyant force acting on the ice cube is given by the formula:
Buoyant force = weight of water displaced = ρ_water * g * V_displaced
where g is the acceleration due to gravity, and V_displaced is the volume of water displaced by the ice cube.
The weight of the ice cube is given by the formula:
Weight of ice cube = ρ_ice * g * V_ice
where V_ice is the volume of the ice cube.
Since the ice cube is in equilibrium (not sinking or rising), the buoyant force and the weight of the ice cube are equal:
ρ_water * g * V_displaced = ρ_ice * g * V_ice
We know that the volume submerged in the water is 1/9 of the volume of the ice cube, so we can express V_displaced as (1/9) * V_ice:
ρ_water * g * (1/9) * V_ice = ρ_ice * g * V_ice
We can simplify the equation by canceling out g and V_ice:
ρ_water * (1/9) = ρ_ice
Therefore, the density of the ice cube is 1/9 times the density of water.
Keep in mind that the density of water is approximately 1000 kg/m^3. Therefore, the density of the ice cube is:
ρ_ice = (1/9) * 1000 kg/m^3
So, the density of ice is approximately 111.11 kg/m^3.