Physics · Mechanical Properties of Matter

JEE Advanced 2024 — Paper 2 — Question 24

A spherical soap bubble inside an air chamber at pressure $P_{0}=10^{5} \mathrm{~Pa}$ has a certain radius so that the excess pressure inside the bubble is $\Delta \mathrm{P}=144 \mathrm{~Pa}$ . Now, the chamber pressure is reduced to $8 \mathrm{P}_{0} / 27$ 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 $\Delta P$ in both the cases to be much smaller than the chamber pressure. The new excess pressure $\Delta P$ in Pa is

Answer: 96

Numerical answer — enter this value.

Step-by-step solution

Since 144 Pa is negligible compared to $\mathrm{P}_{0}$ . Therefore $\mathrm{P}_{0} \frac{4}{3} \pi r_{1}^{3}=\frac{8 \mathrm{P}_{0}}{27}\left(\frac{4}{3} \pi r_{2}^{3}\right)$ $\Rightarrow r_{2}=\frac{3}{2} r_{1}$ $\Rightarrow \Delta \mathrm{P}^{\prime}=\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}=\frac{4 \mathrm{~T}}{\frac{3}{2} \mathrm{r}_{1}}=\frac{\Delta \mathrm{P}_{0} \times 2}{3}=96 \mathrm{~Pa}$

Answer key and solution verified before publishing.

Practise Mechanical Properties of Matter

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Advanced 2024
Paper: Paper 2
Subject: Physics
Chapter: Mechanical Properties of Matter
Topic: Surface Tension and Surface Energy