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) [22 Jan 2026 Shift 1]

Correct Answer is : PB=PAexp(Mω2l2∕2RT)

P_B=P_A \exp (M \omega^2 l^2 \/ 3 R T)

P_B=P_A \exp (M \omega^2 l^2 \/ 2 R T)

P_B=P_A \exp (M \omega^2 l^2 \/ R T)

JEE Main 22 Jan 2026 Shift 1 Paper Solved Paper

Section: Physics

Question : 28 of 75

Marks: +1, -0

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 PA and PB are the pressures at A and B respectively, then (Consider the temperature is same at all points in the tube)

[22 Jan 2026 Shift 1]

PB=PA‌exp(Mω2l2∕3RT)
PB=PA
PB=PA‌exp(Mω2l2∕2RT)
PB=PA‌exp(Mω2l2∕RT)

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