A wooden cylinder floats vertically in water | vedclass

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(B) For a floating object,the weight of the object is equal to the weight of the displaced fluid (Archimedes' Principle).Let $V$ be the total volume o

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Half the density of water

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(B) For a floating object,the weight of the object is equal to the weight of the displaced fluid (Archimedes' Principle).Let $V$ be the total volume of the cylinder,$\rho$ be the density of the wood,and $\sigma$ be the density of water.The weight of the cylinder is $W = V \rho g$.The volume of water displaced is $V_{disp} = \frac{V}{2}$.The buoyant force is $F_B = V_{disp} \sigma g = \frac{V}{2} \sigma g$.Since the cylinder is floating,$W = F_B$.$V \rho g = \frac{V}{2} \sigma g$.Therefore,$\rho = \frac{\sigma}{2}$.This means the density of wood is half the density of water.

$A$ wooden cylinder floats vertically in water with half of its length immersed. The density of wood is

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