AktiraAktira JEE
PhysicsJEE Main 22 Jan 2026, Shift 1·Q12·MCQ·Tough

Mechanical Properties of Fluids · Fluid statics

A cylindrical tube ABAB of length ll, closed at both ends contains an ideal gas of 1 mol having molecular weight MM. The tube is rotated in a horizontal plane with constant angular velocity ω\omega about an axis perpendicular to ABAB and passing through the edge at end A, as shown in the figure. If PAP_A and PBP_B are the pressures at AA and BB respectively, then (Consider the temperature is same at all points in the tube)
  1. (1)PB=PAexp⁡(Mω2l2/2RT)P_B = P_A\exp\left(M\omega^2l^2/2RT\right)
  2. (2)PB=PAP_B = P_A
  3. (3)PB=PAexp⁡(Mω2l2/3RT)P_B = P_A\exp\left(M\omega^2l^2/3RT\right)
  4. (4)PB=PAexp⁡(Mω2l2/RT)P_B = P_A\exp\left(M\omega^2l^2/RT\right)

Video solution coming soon

We're animating every PYQ step by step. Subscribe to get them first.

Subscribe

See the full 22 Jan 2026, Shift 1 paper

More on Fluid statics