Physics · Kinetic Theory of Gases

JEE Advanced 2021 — Paper 2 — Question 27

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 5 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=3.3cc\mathrm{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 p0+Δpp_{0}+\Delta p without changing its orientation. At this pressure,

the volume of the trapped air is v0−Δv\mathrm{v}_{0}-\Delta \mathrm{v}. Let Δv=X\Delta \mathrm{v}=\mathrm{X} cc and

Δp=Y×103 Pa\Delta \mathrm{p}=\mathrm{Y} \times 10^{3} \mathrm{~Pa}.

The value of Y is   ⁣ ⁣  ⁣ ⁣ \text{ }\!\!~\!\!\text{ } .

Answer: 10.00

Numerical answer — enter this value.

Step-by-step solution

For isothermal process

\begin{array}{*{35}{r}}{} & \left( {{10}^{5}} \right)\left( 3.3 \right)=\left( \text{P} \right)\left( 3 \right) \\{} & \text{ }\!\!~\!\!\text{ So, }\!\!~\!\!\text{ P}=1.1\times {{10}^{5}}\text{ }\!\!~\!\!\text{ Pa} \\{} & \text{ }\!\!~\!\!\text{ So, }\!\!~\!\!\text{ }\!\!\Delta\!\!\text{ P}=\left( 1.1-1 \right)\times {{10}^{5}}=10\times {{10}^{3}}\text{ }\!\!~\!\!\text{ Pa} \\\end{array}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2021
Paper
Paper 2
Subject
Physics
Chapter
Kinetic Theory of Gases
Topic
Energy of Gas and Gas Laws and Miscellaneous Problems
A Soft plastic bottle, filled with water of density 1 gm / cc … | JEE Advanced 2021 PYQ with Solution · DhiX AI