A solid weighs 10.5N in air and weighs 55N when immersnd in a liquid of density 900kgm-3. calculate the volume of the solid.?
something is wrong here. How can a solid weigh more when immersed in a fluid?
To calculate the volume of the solid, we can use Archimedes' principle. According to Archimedes' principle, the buoyant force acting on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.
In this case, the weight of the solid in air is 10.5N. When the solid is submerged in the liquid, its weight increases to 55N. The difference in weight between the two situations is due to the buoyant force.
The buoyant force is equal to the weight of the liquid displaced by the solid. Therefore, we can calculate the buoyant force using the density of the liquid and the volume of the solid.
The formula for the buoyant force (FB) is:
FB = ρVg
Where:
ρ is the density of the liquid
V is the volume of the solid
g is the acceleration due to gravity (approximately 9.8 m/s^2)
In this case, the density of the liquid is 900 kg/m^3, and the buoyant force is the difference in weight, which is 55N - 10.5N = 44.5N.
So, we have the equation:
44.5N = 900 kg/m^3 * V * 9.8 m/s^2
To isolate V, we divide both sides of the equation by ρg:
V = FB / (ρg)
Substituting the known values, we get:
V = 44.5N / (900 kg/m^3 * 9.8 m/s^2)
V ≈ 0.005 m^3
Therefore, the volume of the solid is approximately 0.005 m^3.