Calculate Volume Reduction of Trapped Air for Sinking Test Tube

To find the value of $X$, we analyze the condition under which the test tube starts to sink.

1. Volume of the Glass Test Tube

The mass of the glass test tube is $m = 5\text{ g}$ and its density is $\rho_{\text{glass}} = 2.5\text{ g/cc}$.
The volume occupied by the glass material itself is: $V_{\text{glass}} = \frac{m}{\rho_{\text{glass}}} = \frac{5\text{ g}}{2.5\text{ g/cc}} = 2\text{ cc}$

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2. Condition for Sinking

The inverted test tube system (consisting of the glass and the trapped air) floats due to the buoyant force exerted by the water. Sinking begins when the total upward buoyant force on the system becomes less than or equal to the gravitational force acting on the system (neglecting the negligible mass of the trapped air):

$F_{\text{buoyant}} = W_{\text{system}}$

The total volume of water displaced by the test tube system is the sum of the volume of the glass material and the volume of the trapped air, $v$: $V_{\text{displaced}} = V_{\text{glass}} + v$

Thus, the condition for the test tube to just start sinking is: $(V_{\text{glass}} + v) \cdot \rho_{\text{water}} \cdot g = m \cdot g$

Given that the density of water is $\rho_{\text{water}} = 1\text{ g/cc}$: $(2 + v) \times 1 = 5$ $2 + v = 5 \implies v = 3\text{ cc}$

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3. Calculating $X$

The initial volume of the trapped air is $v_0 = 3.3\text{ cc}$. At the verge of sinking, the volume of trapped air decreases to $v_0 - \Delta v = 3\text{ cc}$.

Therefore, the reduction in volume $\Delta v$ is: $\Delta v = v_0 - v = 3.3\text{ cc} - 3.0\text{ cc} = 0.3\text{ cc}$

Given that $\Delta v = X\text{ cc}$, we find: $X = 0.3$