a piece of wood floats on liquid with 4/5 of its volume submerged.if the density of wood is 800 kg/m3 then the density of liquid is?
Dw/Dl = 4/5
800/Dl = 4/5
4Dl = 4000
Dl = 1000 kg/m^3. = Density of the fluid.
1000kg/m3
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To find the density of the liquid, we can use the principle of buoyancy. The buoyant force acting on an object floating in a liquid is equal to the weight of the liquid displaced by the object. The weight of the liquid displaced can be calculated by multiplying the density of the liquid by the volume of the liquid displaced.
Given information:
Density of wood (ρwood) = 800 kg/m^3
Volume submerged (Vsubmerged) = 4/5 of the volume of the wood
Let's assume the volume of the wood is Vwood. Therefore, the volume of the liquid displaced is Vdisp = Vsubmerged = 4/5 * Vwood.
The buoyant force acting on the wood is equal to the weight of the liquid displaced:
Buoyant force = Weight of the liquid displaced
Weight of the liquid displaced = density of liquid (ρliquid) * Vdisp
Since the wood is floating, the buoyant force is equal to the weight of the wood:
Buoyant force = Weight of the wood
Weight of the wood = density of wood (ρwood) * Vwood
Since the buoyant force equals the weight of the wood:
density of liquid (ρliquid) * Vdisp = density of wood (ρwood) * Vwood
Substituting the values we know:
ρliquid * (4/5 * Vwood) = 800 kg/m^3 * Vwood
Simplifying the equation:
4/5 * ρliquid = 800 kg/m^3
To find the density of the liquid (ρliquid):
Multiply both sides of the equation by 5/4:
ρliquid = (5/4) * 800 kg/m^3
ρliquid = 1000 kg/m^3
Therefore, the density of the liquid is 1000 kg/m^3.