1. Given data

• Density of water: $\rho_w = 1\,\text{gm/cc}$
• Mass of test tube: $m = 5\,\text{gm}$
• Density of glass: $\rho_g = 2.5\,\text{gm/cc}$
• Initial pressure: $p_0 = 10^5\,\text{Pa}$
• Initial trapped air volume: $v_0 = 3.3\,\text{cc}$
• At sinking point:
  • pressure $= p_0 + \Delta p$
  • air volume $= v_0 - \Delta v$

We need $\Delta p = Y\times 10^3\,\text{Pa}$.

2. Condition for the test tube to just begin sinking

At the threshold of sinking, the test tube is just neutrally buoyant.

So, $\text{buoyant force} = \text{weight}$

Since density of water is $1\,\text{gm/cc}$, the volume of displaced water (in cc) must equal the mass (in gm), i.e. $V_{\text{displaced}} = 5\,\text{cc}$

3. Volume of glass part of the test tube

Given density of glass $=2.5\,\text{gm/cc}$, so glass volume is $V_g = \frac{m}{\rho_g} = \frac{5}{2.5} = 2\,\text{cc}$

Thus, when it is neutrally buoyant, total displaced volume must be $5\,\text{cc}$. This displaced volume consists of:

• glass volume = $2\,\text{cc}$
• trapped air volume = volume inside submerged tube

Hence trapped air volume at threshold is $V_{air} = 5 - 2 = 3\,\text{cc}$

Therefore, $v_0 - \Delta v = 3\,\text{cc}$

Since $v_0=3.3\,\text{cc}$, $\Delta v = 3.3 - 3 = 0.3\,\text{cc}$

So, $X=0.3$

4. Use Boyle's law (isothermal compression)

Temperature is constant, so for trapped air: $p_0 v_0 = (p_0+\Delta p)(v_0-\Delta v)$

Substitute values: $10^5 \times 3.3 = (10^5+\Delta p)\times 3.0$

Divide by 3: $10^5 \times 1.1 = 10^5 + \Delta p$

So, $1.1\times 10^5 = 10^5 + \Delta p$ $\Delta p = 0.1\times 10^5 = 10^4\,\text{Pa}$

Thus, $\Delta p = 10\times 10^3\,\text{Pa}$

Hence, $Y=10$

5. Comparison with stored answer

Stored correct answer: $10$

Our derived answer is also $10$.