A cylindrical tube A B of length l , closed at both ends contains an ideal gas of 1 mol having molecular weight M . The tube is rotated in a horizontal plane with constant angular velocity ω about an axis perpendicular to A B and passing through the edge at end A , as shown in the figure. If P A and P B are the pressures at A and B respectively, then (Consider the temperature is same at all points in the tube)

P B =P A exp (M ω 2 l 2 / 2 R T )

P B =P A exp (M ω 2 l 2 / 3 R T )

P B =P A exp (M ω 2 l 2 / R T )

Physics · Thermodynamics

JEE Main 2026 — 22 January, Morning Shift — Question 35

A cylindrical tube $A B$ of length $l$ , closed at both ends contains an ideal gas of 1 mol having molecular weight $M$ . The tube is rotated in a horizontal plane with constant angular velocity $\omega$ about an axis perpendicular to $A B$ and passing through the edge at end A , as shown in the figure. If $P_{A}$ and $P_{B}$ are the pressures at $A$ and $B$ respectively, then (Consider the temperature is same at all points in the tube)

Option A:
$P_{B}=P_{A} \exp \left(M \omega^{2} l^{2} / 2 R T\right)$
Correct
Option B:
$P_{B}=P_{A}$
Option C:
$P_{B}=P_{A} \exp \left(M \omega^{2} l^{2} / 3 R T\right)$
Option D:
$P_{B}=P_{A} \exp \left(M \omega^{2} l^{2} / R T\right)$

Answer: A

Step-by-step solution

$\mathrm{A}[(\mathrm{P}+\mathrm{dP})-\mathrm{P}]=(\mathrm{dm})\left(\omega^{2} \mathrm{x}\right)$ $\mathrm{dP}=\frac{(\mathrm{dm})}{\mathrm{A}} \omega^{2} \mathrm{x}$ $\mathrm{dP}=\frac{(\rho)(\mathrm{A})(\mathrm{dx}) \omega^{2} \mathrm{x}}{\mathrm{A}}$ also $[\mathrm{PM}=\rho \mathrm{RT}]$ $\rho=\frac{\mathrm{PM}}{\mathrm{RT}}$ $\mathrm{dP}=\left(\frac{\mathrm{PM}}{\mathrm{RT}}\right) \omega^{2} \mathrm{xdx}$ $\int_{\mathrm{P}_{\mathrm{A}}}^{\mathrm{P}_{\mathrm{B}}} \frac{\mathrm{dP}}{\mathrm{P}}=\frac{\omega^{2} \mathrm{M}}{\mathrm{RT}} \int_{0}^{\ell} \mathrm{xdx}$ $\ell \mathrm{n}\left(\frac{\mathrm{P}_{\mathrm{B}}}{\mathrm{P}_{\mathrm{A}}}\right)=\frac{\omega^{2} \ell^{2} \mathrm{M}}{2 \mathrm{RT}}$ $P_{B}=P_{A} e^{\frac{M \omega^{2} \ell^{2}}{2 R T}}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 22 January, Morning Shift
Subject: Physics
Chapter: Thermodynamics
Topic: Different Thermodynamic Processes

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Answer: A

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