A soft plastic bottle, filled with water of density 1 gm/cc, carries an inverted glass test-tube with some air (ideal gas) trapped as shown in the figure. The test-tube has a mass of 5 gm, and it is made of a thick glass of density 2.5 gm/cc. Initially the bottle is sealed at atmosphere pressure p0 = 105 Pa so that the volume of the trapped air is v0 = 3.3 cc. When the bottle is squeezed from outside at constant temperature, the pressure inside rises and the volume of the trapped air reduces. It is found that the test tube begins to sink at pressure p0 + \Delta p without changing its orientation. At this pressure, the volume of the trapped air is v0 -\Delta v. Let \Delta v = X cc and \Delta p = Y \times 103 Pa. The value of X is \underline\hspace2cm .

Heat and Thermodynamics PYQ — JEE Advanced 2021 · Shift 2 · Q45 with Solution

View written solutionFree

Correct answer: 0.3

Given data

Density of water: $\rho_w = 1\,\text{g/cc}$
Mass of test tube: $m = 5\,\text{g}$
Density of glass: $\rho_g = 2.5\,\text{g/cc}$
Initial air volume trapped in tube: $v_0 = 3.3\,\text{cc}$
Pressure changes isothermally.

We need the decrease in trapped air volume $\Delta v = X\,\text{cc}$ when the test tube just begins to sink.

Condition for just beginning to sink

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

So,

\text{Buoyant force} = \text{Weight of test tube}

Since everything is in cgs units conveniently, buoyancy in terms of displaced water mass is:

\rho_w V_{\text{displaced}} = m

Because $\rho_w = 1\,\text{g/cc}$ ,

V_{\text{displaced}} = 5\,\text{cc}

What volume is displacing water?

The submerged object is the glass test tube plus the trapped air inside it.

So displaced volume at threshold is

V_{\text{glass}} + V_{\text{air}} = 5\,\text{cc}

Now volume of glass is

V_{\text{glass}} = \frac{m}{\rho_g} = \frac{5}{2.5} = 2\,\text{cc}

Hence at the sinking threshold,

2 + V_{\text{air}} = 5

V_{\text{air}} = 3\,\text{cc}

Find decrease in air volume

Initially,

v_0 = 3.3\,\text{cc}

At threshold,

v_0 - \Delta v = 3.0\,\text{cc}

So,

\Delta v = 3.3 - 3.0 = 0.3\,\text{cc}

Thus,

X = 0.3

Comparison with stored correct answer

Stored correct answer = $0.3$

Our derived answer matches it.

More from Heat and Thermodynamics

View written solutionFree

Correct answer: 0.3

Given data

Density of water: $\rho_w = 1\,\text{g/cc}$
Mass of test tube: $m = 5\,\text{g}$
Density of glass: $\rho_g = 2.5\,\text{g/cc}$
Initial air volume trapped in tube: $v_0 = 3.3\,\text{cc}$
Pressure changes isothermally.

We need the decrease in trapped air volume $\Delta v = X\,\text{cc}$ when the test tube just begins to sink.

Condition for just beginning to sink

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

So,

\text{Buoyant force} = \text{Weight of test tube}

Since everything is in cgs units conveniently, buoyancy in terms of displaced water mass is:

\rho_w V_{\text{displaced}} = m

Because $\rho_w = 1\,\text{g/cc}$ ,

V_{\text{displaced}} = 5\,\text{cc}

What volume is displacing water?

The submerged object is the glass test tube plus the trapped air inside it.

So displaced volume at threshold is

V_{\text{glass}} + V_{\text{air}} = 5\,\text{cc}

Now volume of glass is

V_{\text{glass}} = \frac{m}{\rho_g} = \frac{5}{2.5} = 2\,\text{cc}

Hence at the sinking threshold,

2 + V_{\text{air}} = 5

V_{\text{air}} = 3\,\text{cc}

Find decrease in air volume

Initially,

v_0 = 3.3\,\text{cc}

At threshold,

v_0 - \Delta v = 3.0\,\text{cc}

So,

\Delta v = 3.3 - 3.0 = 0.3\,\text{cc}

Thus,

X = 0.3

Comparison with stored correct answer

Stored correct answer = $0.3$

Our derived answer matches it.

Heat and Thermodynamics question

Heat and Thermodynamics question

More from Heat and Thermodynamics

Heat and Thermodynamics question

Heat and Thermodynamics question

More from Heat and Thermodynamics