A cylindrical tube A B of… - JEE Main 2026 Physics PYQ

JEE Main 2026 — Physics Thermodynamics

2026mcqhard

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)

JEE Main 2026 Physics, Kinetic Theory of Gases — question figure

Official previous-year question

Held on 22 Jan 2026 · Verified 6 Jul 2026.

Options

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

Did you get this right?

Solution

JEE Main 2026 Physics, Kinetic Theory of Gases — solution figure

$A[(P + dP) - P] = (dm)(\omega^2 x)$

$dP = \dfrac{(dm)}{A} \omega^2 x$

$dP = \dfrac{(\rho)(A)(dx)\omega^2 x}{A}$

also $[PM = \rho RT]$

$\rho = \dfrac{PM}{RT}$

$dP = \left( \dfrac{PM}{RT} \right) \omega^2 x \, dx$

$\int_{P_A}^{P_B} \dfrac{dP}{P} = \dfrac{\omega^2 M}{RT} \int_{0}^{\ell} x \, dx$

$\ln \left( \dfrac{P_B}{P_A} \right) = \dfrac{\omega^2 \ell^2 M}{2RT}$

$P_B = P_A e^{\dfrac{M\omega^2 \ell^2}{2RT}}$

Your note

JEE Main Thermodynamics in other years

Thermodynamics 2025 Thermodynamics 2024 Thermodynamics 2023 Thermodynamics 2022 Thermodynamics 2021 Thermodynamics 2020 Thermodynamics 2019 Thermodynamics 2018 Thermodynamics 2017 Thermodynamics 2016 Thermodynamics 2015 Thermodynamics 2014

Keep practicing Thermodynamics

Work through every JEE Main Thermodynamics PYQ, year by year.

Browse all Physics topics

Options

$P_{B}=P_{A}$

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

$P_{B}=P_{A} \exp \left(M \omega^{2} l^{2} / R T\right)$

$P_{B}=P_{A} \exp \left(M \omega^{2} l^{2} / 2 R T\right)$

Solution

JEE Main 2026 Physics, Kinetic Theory of Gases — solution figure

$A[(P + dP) - P] = (dm)(\omega^2 x)$

$dP = \dfrac{(dm)}{A} \omega^2 x$

$dP = \dfrac{(\rho)(A)(dx)\omega^2 x}{A}$

also $[PM = \rho RT]$

$\rho = \dfrac{PM}{RT}$

$dP = \left( \dfrac{PM}{RT} \right) \omega^2 x \, dx$

$\int_{P_A}^{P_B} \dfrac{dP}{P} = \dfrac{\omega^2 M}{RT} \int_{0}^{\ell} x \, dx$

$\ln \left( \dfrac{P_B}{P_A} \right) = \dfrac{\omega^2 \ell^2 M}{2RT}$

$P_B = P_A e^{\dfrac{M\omega^2 \ell^2}{2RT}}$

Options

Did you get this right?

Solution

Your note

JEE Main Thermodynamics in other years

More JEE Main Thermodynamics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Optics

Waves & Oscillations

Keep practicing Thermodynamics

Options

Did you get this right?

Solution

Your note

JEE Main Thermodynamics in other years

More JEE Main Thermodynamics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Optics

Waves & Oscillations

Keep practicing Thermodynamics