A rectangular block of wood floats in water with two-third of it's volume immersed .when placed placed in another liquid ,it floats with half of it's volume immersed .calculate the relative density of a liduid.
2/3 * Dw = 1/2 * D?
I have no idea
To calculate the relative density of a liquid in this scenario, we need to use the principle of buoyancy.
Let's denote the density of water as ρ_water, the density of the block as ρ_block, and the density of the unknown liquid as ρ_liquid.
According to the information given:
1) In water, two-thirds of the volume of the block is immersed. This means the immersed volume is (2/3) times the total volume of the block.
2) In the unknown liquid, half of the volume of the block is immersed. This means the immersed volume is (1/2) times the total volume of the block.
Using these observations and the principle of buoyancy (which states that the weight of the liquid displaced by the object is equal to the weight of the object itself), we can set up the following equations:
For the block in water:
ρ_block * (2/3) * V_block * g = ρ_water * (2/3) * V_block * g
Here, V_block represents the total volume of the block and g is the acceleration due to gravity.
For the block in the unknown liquid:
ρ_block * (1/2) * V_block * g = ρ_liquid * (1/2) * V_block * g
Notice that the volume of the block cancels out in both equations, simplifying the equations to:
ρ_block = ρ_water
ρ_block = ρ_liquid
Therefore, the relative density of the liquid (ρ_liquid) is equal to the density of water (ρ_water). The relative density is a dimensionless quantity, so it does not have any units.
In conclusion, the relative density of the liquid is the same as that of water.