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QJEEAdv21Phc42f: JEE Advanced 2021 Physics - Class 11 JEEAdv21Phc42f Solution: A soft plastic bottle, filled with water of density 1 gm/cc,…

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 + Δ\Delta p without changing its orientation. At this pressure, the volume of the trapped air is v0 −- Δ\Delta v. Let Δ\Delta v = X cc and Δ\Delta p = Y ×\times 103 Pa.

JEE Advanced 2021 Paper 2 Online Physics - Heat and Thermodynamics Question 32 English

The value of Y is _______________.

Step-by-Step Verified Solution

When the tube just sinks/floats, then average density = density of water

Masstotal volume{{Mass} \over {total\,volume}} = 1 gm/cc

⇒5gmtotal volume\Rightarrow {{5gm} \over {total\,volume}} = 1 gm/cc

⇒\Rightarrow Total volume = 5 cc

⇒\Rightarrow Volume of tube + final volume of air in the tube = 5 cc

⇒5gm2.5gm/cc+Vf=5\Rightarrow {{5gm} \over {2.5gm/cc}} + {V_f} = 5

⇒\Rightarrow Vf = 5 −- 2 = 3 cc

⇒\Rightarrow Δ\Delta V = 0.3 cc

For isothermal process,

piVi=pfVf⇒pf=105×3.33{p_i}{V_i} = {p_f}{V_f} \Rightarrow {p_f} = {10^5} \times {{3.3} \over 3}

pf = 1.1 ×\times 105

pf −- pi = 1.1 ×\times 105 −- 105

= 0.1 ×\times 105 = 10 ×\times 103 Pa ⇒\Rightarrow Y = 10

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