A cylindrical tube $AB$ 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 $AB$ 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)

(A) $P_B = P_A \exp(M\omega^2 l^2 / 2RT)$

(C) $P_B = P_A \exp(M\omega^2 l^2 / 3RT)$

(D) $P_B = P_A \exp(M\omega^2 l^2 / RT)$

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Heat & Thermodynamics

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)

Answer: (A)

Take a gas element of thickness

d x

at distance

x

from the axis through

A

. The pressure difference across it provides the centripetal force,

d P = ρ ω^{2} x d x

, and with

ρ = \frac{P M}{R T}

this becomes

\frac{d P}{P} = \frac{M ω ^{2}}{R T} x d x

. Integrating from

A

(

x = 0

) to

B

(

x = l

ln \frac{P _{B}}{P _{A}} = \frac{M ω ^{2} l ^{2}}{2 R T}

, so

P_{B} = P_{A} exp (\frac{M ω ^{2} l ^{2}}{2 R T}) .

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Single correct option