The value of Y is ___ .

The mass of the glass test-tube is $$m = 5\,$$g and the density of glass is $$\rho_g = 2.5\,$$g cc-1.

Volume of the glass material:
$$V_g = \frac{m}{\rho_g} = \frac{5}{2.5} = 2 \text{ cc}$$

The trapped air initially occupies $$v_0 = 3.3 \text{ cc}$$. Therefore, at the start, the total volume that excludes water (and hence displaces water) is
$$V_{\text{disp,0}} = V_g + v_0 = 2 + 3.3 = 5.3 \text{ cc}$$

Bouyant force equals weight of the water of this displaced volume. When the test-tube just begins to sink, the buoyant force must equal its own weight (air’s mass is negligible):
$$\rho_{\text{water}} g (V_g + v_0 - \Delta v) = mg$$

Taking $$\rho_{\text{water}} = 1 \text{ g\,cc}^{-1}$$ and cancelling $$g$$ gives
$$V_g + v_0 - \Delta v = m \quad -(1)$$

Substituting the numbers into (1):
$$2 + 3.3 - \Delta v = 5$$
$$\Rightarrow \Delta v = 0.3 \text{ cc}$$

The trapped air behaves as an ideal gas at constant temperature, so
$$p_0 v_0 = (p_0 + \Delta p)\,(v_0 - \Delta v) \quad -(2)$$

With $$p_0 = 10^5 \text{ Pa}$$, $$v_0 = 3.3 \text{ cc}$$ and $$\Delta v = 0.3 \text{ cc}$$, equation (2) becomes
$$10^5 \times 3.3 = (10^5 + \Delta p)\times 3.0$$

Simplifying:
$$10^5 + \Delta p = 10^5 \times \frac{3.3}{3.0} = 1.1 \times 10^5$$
$$\Rightarrow \Delta p = (1.1 - 1.0)\times 10^5 = 0.1 \times 10^5 \text{ Pa} = 1.0 \times 10^4 \text{ Pa}$$

Since $$\Delta p = Y \times 10^3 \text{ Pa}$$,
$$Y = \frac{1.0 \times 10^4}{10^3} = 10$$

Hence, $$Y \approx 10$$, which lies in the accepted range 9.00 - 10.10.

Answer: 10