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JEE Main 2026 (22 Jan, Shift 1) · Rotational Kinematics & Dynamics

Rotational Kinematics & Dynamics question from JEE Main 2026 (22 Jan, Shift 1)

Q12JEE Main 2026 · 22 JAN · SHIFT 1Difficult

Rotational Kinematics & Dynamics

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 AA, 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)

Figure

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Answer: (c)

  1. Pressure gradient gives centripetal force: dPdx=ρω2x\dfrac{dP}{dx} = \rho\omega^2 x, with ρ=PMRT\rho = \dfrac{PM}{RT}.
  2. ∫PAPBdPP=Mω2RT∫0lx dx⇒ln⁡PBPA=Mω2l22RT\int_{P_A}^{P_B}\dfrac{dP}{P} = \dfrac{M\omega^2}{RT}\int_0^l x\,dx \Rightarrow \ln\dfrac{P_B}{P_A} = \dfrac{M\omega^2 l^2}{2RT}
  3. PB=PAeMω2l2/2RTP_B = P_A e^{M\omega^2 l^2/2RT}, so the answer is (c).
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