A piece of wood of mass 40g and uniform cross-sectional area of 2cm2 floats upright in water. The length of the wood immersed is
Question 1 Report
A piece of wood of mass 40g and uniform cross-sectional area of 2cm2 floats upright in water. The length of the wood immersed is
Answer Details
To float upright, the buoyant force on the wood must be equal to the weight of the wood. The buoyant force is equal to the weight of the water displaced by the wood, which is also equal to the volume of the water displaced multiplied by the density of water. Since the wood floats upright, the volume of the water displaced is equal to the volume of the part of the wood that is submerged.
Let's assume that the length of the wood immersed is "x" cm. Then the volume of the water displaced is the cross-sectional area of the wood (2 cm²) multiplied by the length of the wood immersed (x cm). Therefore, the buoyant force on the wood is:
Buoyant force = Volume of water displaced × Density of water
Buoyant force = (2 cm² × x cm) × 1 g/cm³
Buoyant force = 2x g
The weight of the wood is 40 g. For the wood to float upright, the buoyant force must be equal to the weight of the wood:
2x g = 40 g
Solving for x, we get:
x = 20 cm
Therefore, the length of the wood immersed is 20 cm.
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