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JEE Advanced 2024Paper 2
Exam
JEE Advanced
Year
2024
Paper
Paper 2
Subject
Physics
Chapter
Mechanical Properties of Fluids

JEE Advanced 2024 · Paper 2 · Physics · Mechanical Properties of Fluids

A spherical soap bubble inside an air chamber at pressure P_0=10^5...

A spherical soap bubble inside an air chamber at pressure P0=105 PaP_0=10^5 \mathrm{~Pa} has a certain radius so that the excess pressure inside the bubble is ΔP=144 Pa\Delta P=144 \mathrm{~Pa}. Now, the chamber pressure is reduced to 8P0/278 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 ΔP\Delta P in both the cases to be much smaller than the chamber pressure. The new excess pressure ΔP\Delta P in Pa\mathrm{Pa} is ________ .

Correct answer

0

Step-by-step solution

Case-1 P−P0=ΔP=4TRP=(P0+4TR)\begin{aligned} & \mathrm{P}-\mathrm{P}_0=\Delta \mathrm{P}=\frac{4 \mathrm{ T}}{\mathrm{R}} \\ & \mathrm{P}=\left(\mathrm{P}_0+\frac{4 \mathrm{ T}}{\mathrm{R}}\right)\end{aligned} Case-2 P1−8P027=ΔP1=4TR1P1=4TR1+8P027\begin{aligned} & \mathrm{P}_1-\frac{8 \mathrm{P}_0}{27}=\Delta \mathrm{P}_1=\frac{4 \mathrm{ T}}{\mathrm{R}_1} \\ & \mathrm{P}_1=\frac{4 \mathrm{ T}}{\mathrm{R}_1}+\frac{8 \mathrm{P}_0}{27}\end{aligned} Constant temperature process PV=P1V1(P0+4TR)43πR3=(4TR1+8P027)43πR13;(4TR),(4TR1)→ (Neglected) R=23R1⇒R1=32RΔP1=4TR1=4T3R×2=23×(144)=96Pa\begin{aligned} & \mathrm{PV}=\mathrm{P}_1 \mathrm{ V}_1 \\ & \left(\mathrm{P}_0+\frac{4 \mathrm{ T}}{\mathrm{R}}\right) \frac{4}{3} \pi \mathrm{R}^3=\left(\frac{4 \mathrm{ T}}{\mathrm{R}_1}+\frac{8 \mathrm{P}_0}{27}\right) \frac{4}{3} \pi \mathrm{R}_1^3 ;\left(\frac{4 \mathrm{ T}}{\mathrm{R}}\right),\left(\frac{4 \mathrm{ T}}{\mathrm{R}_1}\right) \rightarrow \text{ (Neglected) } \\ & \mathrm{R}=\frac{2}{3} \mathrm{R}_1 \Rightarrow \mathrm{R}_1=\frac{3}{2} \mathrm{R} \\ & \Delta \mathrm{P}_1=\frac{4 \mathrm{ T}}{\mathrm{R}_1}=\frac{4 \mathrm{ T}}{3 \mathrm{R}} \times 2=\frac{2}{3} \times(144)=96 \mathrm{ Pa}\end{aligned}
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