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

The correct answer is Half the density of water.

To understand why this is the case, we can apply the principles of buoyancy and flotation (Archimedes' principle).

Step 1: Define the variables
Let the total length of the wooden cylinder be $L$ and its cross-sectional area be $A$.
Let the density of the wood be ${\rho }_{\text{wood}}$ and the density of water be ${\rho }_{\text{water}}$.
Let $g$ be the acceleration due to gravity.

Step 2: Express the weight of the cylinder
The total volume $V$ of the cylinder is:
$V=A·L$
The weight $W$ of the wooden cylinder is the product of its volume, density, and gravity:
$W=V·{\rho }_{\text{wood}}·g=A·L·{\rho }_{\text{wood}}·g$

Step 3: Express the buoyant force
The cylinder floats vertically with half of its length immersed in water. Therefore, the depth of the submerged portion is $\frac{L}{2}$.
The volume of water displaced by this submerged portion, $V_{\text{displaced}}$, is:
$V_{\text{displaced}}=A·\frac{L}{2}$
According to Archimedes' principle, the buoyant force $F_B$ is equal to the weight of the displaced water:
$F_B=V_{\text{displaced}}·{\rho }_{\text{water}}·g=A·\frac{L}{2}·{\rho }_{\text{water}}·g$

Step 4: Set up the equilibrium condition
For a floating object to remain in equilibrium, its total downward weight must be balanced by the upward buoyant force:
$W=F_B$
Substituting the expressions we derived:
$A·L·{\rho }_{\text{wood}}·g=A·\frac{L}{2}·{\rho }_{\text{water}}·g$

Step 5: Solve for the density of wood
We can cancel the common terms ($A$, $L$, and $g$) on both sides of the equation:
${\rho }_{\text{wood}}=\frac{1}{2}{\rho }_{\text{water}}$

This shows that the density of the wood is exactly half the density of water.