JEE Advanced
Physics
Mechanical Properties of Fluids
2021
JEE Advanced 2021 (Paper 2)
JEE Advanced Physics Question (2021) — Solution
Question
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 =10^ 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 + p without changing its orientation. At this pressure, the volume of the trapped air is v_ 0 - v. Let v=X cc and p=Y 10^ 3 ~Pa . The value of Y is _____.
Step-by-step solution
When all the forces on tube are balanced then the tube will start sinking. F u p w a r d = F d o w n w a r d ρ w V tube   + V air   g = m tube   g 2   cm 3 + V air   g = ( 5   gm ) g V air = 3   cm , while the initial volume of the air was 3 . 3   cm 3 so the decrease in the volume Δ V = 3 . 3 - 3 = 0 . 3   cm 3 V tube   = m tube   ρ tube   = 5   gm 2 . 5   gm   cm - 3 V tube   = 2   cm 3 Since the temperature of the tapped air remains constant, so according to the ideal gas equation, P V = constant Take ln on both side. ⇒ l n P + l n V = l n (constant) Differentiate both sides, d P P + d V V = 0       ∵ d d t constant   = 0 ⇒ Δ P P = - Δ V V ⇒ Δ P 10 5 = - - 0 . 3 3 . 3 ⇒ Δ P = 1 11 × 10 5   Pa = 100 11 × 10 3 ⇒ Δ P = 9 . 09 × 10 3   Pa Compare it with, Δ P = Y × 10 3   Pa So, Y = 9 . 09
Practice more on Quantrex App →
Related: Physics — Mechanical Properties of Fluids · All PYQ Banks