An object floats in a liquid of density 1.2 kg/m3 with one quarter of its volume above the liquid surface. Determine the density of the object
Vb = Do/Dl * Vo.
3/4 = Do/1.2 * 1.
4Do = 3.6,
Do = 0.9kg/m^3. = Density of the object.
To determine the density of the object, we need to use the concept of buoyancy.
Buoyancy is the upward force exerted on an object submerged in a fluid, such as a liquid. According to Archimedes' principle, the buoyant force is equal to the weight of the fluid displaced by the object.
In this case, we know that the object is floating, which means the buoyant force is equal to the weight of the object. We also know that one quarter of the object's volume is above the liquid surface. This means that three-quarters of the object's volume is submerged in the liquid.
Now, let's calculate the buoyant force. The buoyant force, F_b, is given by the formula:
F_b = ρ_fluid * g * V_submerged
where ρ_fluid is the density of the liquid, g is the acceleration due to gravity, and V_submerged is the volume of the object submerged in the liquid.
In this case, since three-quarters of the object's volume is submerged, V_submerged = (3/4) * V_object, where V_object is the total volume of the object.
Now, let's equate the buoyant force to the weight of the object to find the density:
F_b = ρ_fluid * g * V_submerged
Weight = ρ_object * g * V_object
Since the object is floating, the weight of the object is equal to the buoyant force, so we have:
ρ_fluid * g * V_submerged = ρ_object * g * V_object
Simplifying the equation:
ρ_fluid * V_submerged = ρ_object * V_object
Since V_submerged/V_object = 3/4, we can rewrite the equation as:
ρ_fluid * (3/4) * V_object = ρ_object * V_object
Canceling out V_object from both sides:
ρ_fluid * (3/4) = ρ_object
Now, we can plug in the given information. The density of the liquid, ρ_fluid, is 1.2 kg/m^3. Substituting this value into the equation:
1.2 * (3/4) = ρ_object
Multiplying 1.2 by 3/4:
0.9 = ρ_object
Therefore, the density of the object is 0.9 kg/m^3.