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JEE Advanced Physics Mechanical Properties of Fluids 2024 JEE Advanced 2024 (Paper 2)

JEE Advanced Physics Question (2024) — Solution

Question

A spherical soap bubble inside an air chamber at pressure P_0=10^5 ~Pa has a certain radius so that the excess pressure inside the bubble is P=144 ~Pa . Now, the chamber pressure is reduced to 8 P_0 / 27 so that the bubble radius and its excess pressure change. In this process, all the temperatures remain unchanged. Assume air to be an ideal gas and the excess pressure P in both the cases to be much smaller than the chamber pressure. The new excess pressure P in Pa is ________ .

Options

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

Answer

A. A

Step-by-step solution

Case-1 aligned & P - P _0= P = 4 ~T R \\ & P = ( P _0+ 4 ~T R ) aligned Case-2 aligned & P _1- 8 P _0 27 = P _1= 4 ~T R _1 \\ & P _1= 4 ~T R _1 + 8 P _0 27 aligned Constant temperature process aligned & PV = P _1 ~V _1 \\ & ( P _0+ 4 ~T R ) 4 3 R ^3= ( 4 ~T R _1 + 8 P _0 27 ) 4 3 R _1^3 ; ( 4 ~T R ), ( 4 ~T R _1 ) (Neglected) \\ & R = 2 3 R _1 R _1= 3 2 R \\ & P _1= 4 ~T R _1 = 4 ~T 3 R 2= 2 3 (144)=96 ~Pa aligned

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