JEE Main 2026 — Jan 22, Shift 1 — Physics solution

JEE Main2026Jan 22, Shift 1Physics

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 $ω$ 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
$P_{B} = P_{A} exp (\frac{M ω ^{2} l ^{2}}{2 R T})$
B
$P_{B} = P_{A}$
C
$P_{B} = P_{A} exp (\frac{M ω ^{2} l ^{2}}{3 R T})$
D
$P_{B} = P_{A} exp (\frac{M ω ^{2} l ^{2}}{R T})$

Solution

In the rotating frame, a gas element at distance $x$ from the rotation axis feels a centrifugal pseudo-force per unit volume of $ρ ω^{2} x$ , pointing outward.

Force balance on a thin slice of area $A$ , thickness $d x$ , at distance $x$ : $A [(P + d P) - P] = (ρ A d x) ω^{2} x$ , so $d P = ρ ω^{2} x d x$ .

Using ideal-gas $ρ = \frac{P M}{R T}$ : $\frac{d P}{P} = \frac{M ω ^{2}}{R T} x d x$ .

Integrate from A (at $x = 0$ ) to B (at $x = l$ ): $ln \frac{P _{B}}{P _{A}} = \frac{M ω ^{2} l ^{2}}{2 R T}$ .

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

← Back to full paper

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →