A piece of wood is floating in water kept inside a bottle. The...

Let $V$ be the total volume of the wood, $V_{in}$ be the volume submerged in water, $\rho_w$ be the density of water, $\rho_a$ be the density of air, and $\rho$ be the density of wood.

The weight of the wood is balanced by the buoyant force due to water and the buoyant force due to air.

$V \rho g = V_{in} \rho_w g + (V - V_{in}) \rho_a g$

$V_{in} (\rho_w - \rho_a) = V (\rho - \rho_a)$

$V_{in} = V \dfrac{\rho - \rho_a}{\rho_w - \rho_a}$

When more air is pushed into the bottle, the pressure increases, which increases the density of air $\rho_a$.

Differentiating $V_{in}$ with respect to $\rho_a$:

$\dfrac{dV_{in}}{d\rho_a} = V \dfrac{-(\rho_w - \rho_a) - (\rho - \rho_a)(-1)}{(\rho_w - \rho_a)^2} = V \dfrac{\rho - \rho_w}{(\rho_w - \rho_a)^2}$

Since the wood floats in water, $\rho < \rho_w$, which means $\dfrac{dV_{in}}{d\rho_a} < 0$.

Thus, as the density of air increases, the volume of the wood submerged in water ($V_{in}$) decreases. Therefore, it will float with a lesser part in the water.