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 2.5gm/cc. Initially the bottle is sealed at atmospheric pressure p0=105Pa so that the volume of the trapped air is v0=3.3cc. 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+Δp without changing its orientation. At this pressure, the volume of the trapped air is v0−Δv. Let Δv=Xcc and Δp=Y×103Pa.
The value of Y is ____.
Answer: 10
Solution
From the previous question, when the test tube begins to sink, Vgas=3.0 cc.
Applying the isothermal process for the trapped air:
P1V1=P2V2
105×3.3=P2×3.0
P2=3.0105×3.3=1.1×105 Pa
Therefore:
Δp=P2−P1=1.1×105−105=0.1×105=104 Pa
Δp=10×103 Pa
So Y=10.00
JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Numerical answer
