JEE Challenger
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PhyHardAdv 2021 • Paper 2

Determine Pressure Increase Required for Test Tube Sinking

Comprehension Passage

A soft plastic bottle containing water of density 1 gm/cc1\text{ gm/cc} contains an inverted glass test-tube with trapped air (ideal gas) as depicted in the figure. The test-tube mass is 5 gm5\text{ gm} and it is made of thick glass with a density of 2.5 gm/cc2.5\text{ gm/cc}. Initially, the bottle is sealed under atmospheric pressure p0=105 Pap_0 = 10^5\text{ Pa}, with the trapped air occupying a volume of v0=3.3 ccv_0 = 3.3\text{ cc}. When squeezed from the outside under isothermal conditions, the internal pressure increases while the volume of trapped air decreases. It is observed that the test tube starts sinking at a pressure of p0+Δpp_0 + \Delta p maintaining its vertical orientation. Under this pressure, the volume of trapped air becomes v0−Δvv_0 - \Delta v. Let Δv=X cc\Delta v = X\text{ cc} and Δp=Y×103 Pa\Delta p = Y \times 10^3\text{ Pa}.

Calculate the value of YY.

Question Diagram 1
Official Numerical Answer10

Topics & Concepts

Mechanical Properties of FluidsFluid DynamicsKinetic Theory of Gases

Step-by-Step Solution

To determine the value of YY, we analyze the force equilibrium of the test-tube at the moment it begins to sink.

Step 1: Calculate the Volume of the Glass

Given:

  • Mass of the test-tube, m=5 gm = 5\text{ g}
  • Density of the glass, ρg=2.5 g/cc\rho_g = 2.5\text{ g/cc}

The volume occupied by the glass material itself is: Vg=mρg=5 g2.5 g/cc=2 ccV_g = \frac{m}{\rho_g} = \frac{5\text{ g}}{2.5\text{ g/cc}} = 2\text{ cc}

Step 2: Condition for Sinking

The test-tube starts to sink when its average density becomes equal to the density of water (ρw=1 g/cc\rho_w = 1\text{ g/cc}), i.e., when the total weight of the test-tube equals the buoyant force exerted by the displaced water.

Let v′v' be the volume of the trapped air when the test-tube begins to sink. The total volume of water displaced by the test-tube and trapped air is: Vtotal=Vg+v′=(2+v′) ccV_{\text{total}} = V_g + v' = (2 + v')\text{ cc}

For neutral buoyancy: Total Mass=Mass of displaced water\text{Total Mass} = \text{Mass of displaced water} m=ρw×Vtotalm = \rho_w \times V_{\text{total}} 5 g=1 g/cc×(2+v′) cc5\text{ g} = 1\text{ g/cc} \times (2 + v')\text{ cc} v′=3 ccv' = 3\text{ cc}

Step 3: Apply Isothermal Compression (Boyle's Law)

The trapped air undergoes an isothermal process as the internal pressure increases from p0p_0 to p0+Δpp_0 + \Delta p.

Given:

  • Initial atmospheric pressure, p0=105 Pap_0 = 10^5\text{ Pa}
  • Initial air volume, v0=3.3 ccv_0 = 3.3\text{ cc}
  • Final air volume, v′=3.0 ccv' = 3.0\text{ cc}

Using Boyle's law (p1V1=p2V2p_1 V_1 = p_2 V_2): p0v0=(p0+Δp)v′p_0 v_0 = (p_0 + \Delta p) v'

Substitute the given values into the equation: 105×3.3=(105+Δp)×3.010^5 \times 3.3 = (10^5 + \Delta p) \times 3.0 105+Δp=105×3.33.010^5 + \Delta p = 10^5 \times \frac{3.3}{3.0} 105+Δp=1.1×10510^5 + \Delta p = 1.1 \times 10^5 Δp=0.1×105 Pa=104 Pa\Delta p = 0.1 \times 10^5\text{ Pa} = 10^4\text{ Pa}

Step 4: Determine YY

We are given that Δp=Y×103 Pa\Delta p = Y \times 10^3\text{ Pa}. Expressing Δp\Delta p in terms of 103 Pa10^3\text{ Pa}: 104 Pa=10×103 Pa10^4\text{ Pa} = 10 \times 10^3\text{ Pa}

Thus, Y=10Y = 10

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