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JEE Advanced Physics Mechanical Properties of Fluids 2023 JEE Advanced 2023 (Paper 2)

JEE Advanced Physics Question (2023) — Solution

Question

An incompressible liquid is kept in a container having a weightless piston with a hole. A capillary tube of inner radius 0 . 1   mm is dipped vertically into the liquid through the airtight piston hole, as shown in the figure. The air in the container is isothermally compressed from its original volume V 0  to  100 101 V 0  with the movable piston. Considering air as an ideal gas, the height h of the liquid column in the capillary above the liquid level in cm  is _____. [Given: Surface tension of the liquid is 0 . 075   N   m − 1 , atmospheric pressure is 10 5   N   m − 2 , acceleration due to gravity g is 10   m   s − 2 , density of the liquid is 10 3   kg   m − 3  and contact angle of capillary surface with the liquid is zero]

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

Let P f  be the air pressure. As the process is isothermal, we can write P 0 V 0 = P f V f ⇒ P 0 V 0 = P f 100 101 V 0 ⇒ P f = 101 × 10 3   Pa                     ∵   P 0 = 10 5   N   m - 2 Now, consider the 4 points shown in diagram P d - P c = 2 T R                     ∵   P d = P 0 ∴   P c = P 0 - 2 T R Now, P a = P b  (also,  P a = P f ) ⇒ P f = ρ g h + P c ⇒ 101 × 10 3 = 10 3 × 10 × h + 10 5 - 2 × 0 . 075 0 . 1 × 10 - 3 ⇒ h = 1 4   m = 25   cm

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Related: Physics — Mechanical Properties of Fluids · All PYQ Banks