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JEE Advanced 2021Paper 2
Exam
JEE Advanced
Year
2021
Paper
Paper 2
Subject
Physics
Chapter
Mechanical Properties of Fluids

JEE Advanced 2021 · Paper 2 · Physics · Mechanical Properties of Fluids

A soft plastic bottle, filled with water of density 1 gm / cc ,...

A soft plastic bottle, filled with water of density 1gm/cc1 \mathrm{gm} / \mathrm{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 5gm5 \mathrm{gm}, and it is made of a thick glass of density 2.5gm/cc2.5 \mathrm{gm} / \mathrm{cc}. Initially the bottle is sealed at atmospheric pressure p0=105 Pap_{0}=10^{5} \mathrm{~Pa} so that the volume of the trapped air is v0=3.3ccv_{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 p0+Δpp_{0}+\Delta p without changing its orientation. At this pressure, the volume of the trapped air is v0−Δvv_{0}-\Delta v. Let Δv=X\Delta v=X cc and Δp=Y×103 Pa\Delta p=Y \times 10^{3} \mathrm{~Pa}. A soft plastic bottle, filled with water of density 1 gm / cc , carries an inverted glass test-tube with some air The value of XX is ____.

Correct answer

0

Step-by-step solution

When all the forces are balanced on tube, it will start sinking.

Fupward=Fdownward{F}_{upward} = {F}_{downward}

\left({ρ}_{w}\right)\left({V}_{tube} + {V}_{air}\right)g = \left({m}_{tube}\right)g</p><p>(2cm3+Vair)g=(5gm)g</p><p>\left(2{cm}^{3} + {V}_{air}\right)g = ( 5gm ) g

A soft plastic bottle, filled with water of density 1 gm / cc , carries an inverted glass test-tube with some air

Vair=3cm{V}_{air} = 3cm, while the initial volume of the air was 3.3cm33 . 3{cm}^{3} so the decrease in the volume \DeltaV=3.3−3=0.3cm3\DeltaV = 3 . 3 - 3 = 0 . 3{cm}^{3}

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