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 \(2.5 \mathrm{gm} / \mathrm{cc}\). Initially the bottle is sealed at atmospheric pressure \(p_{0}=10^{5} \mathrm{~Pa}\) so that the volume of the trapped air is \(v_{0}=3.3 \mathrm{cc}\). 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 \(P_{0}+\Delta p\) without changing its orientation. At this pressure, the volume of the trapped air is \(v_{0}-\Delta v\). Let \(\Delta v=X \mathrm{cc}\) and \(\Delta p=Y \times 10^{3} \mathrm{~Pa}\). The value of \(X\) is ____.

JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Fluid Mechanics

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 atmospheric pressure

p_{0} = 1 0^{5} Pa

so that the volume of the trapped air is

v_{0} = 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

P_{0} + Δ p

without changing its orientation. At this pressure, the volume of the trapped air is

v_{0} - Δ v

.
Let

Δ v = X cc

and

Δ p = Y \times 1 0^{3} Pa

.

The value of

X

is ____.

Answer: 0.3

Solution

Volume of glass:

V_{glass} = \frac{mass}{density} = \frac{5}{2.5} = 2 cc

Initially, total volume =

V_{glass} + V_{gas} = 2 + 3.3 = 5.3

cc

The test tube begins to sink when buoyant force equals weight:

ρ_{water} (V_{glass} + V_{gas}) g = m g

1 \times (2 + V_{gas}) = 5

V_{gas} = 3 cc

Therefore:

Δ v = v_{0} - V_{gas} = 3.3 - 3.0 = 0.3 cc

X = 0.30

JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Numerical answer

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JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Fluid Mechanics

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