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)
To solve this problem, we need to analyze the effect of rotation on the pressure distribution in the cylindrical tube containing an ideal gas. The tube is rotated with an angular velocity \omega around an axis passing through point A. Let's break down the steps:
Since the centrifugal force does not contribute to any pressure gradient along the length of the tube, the pressure at point B, P_B, equals the pressure at point A, P_A.
Therefore, the correct answer is:

A uniform bar of length 12 cm and mass $20m$ lies on a smooth horizontal table. Two point masses $m$ and $2m$ are moving in opposite directions with same speed of $v$ and in the same plane as the bar, as shown in figure. These masses strike the bar simultaneously and get stuck to it. After collision the entire system is rotating with angular frequency $\omega$. The ratio of $v$ and $\omega$ is :
Match the LIST-I with LIST-II
| List-I: | List-II: |
| A. Magnetic induction | I. $[M L T^{-2} A^{-2}]$ |
| B. Magnetic flux | II. $[M L^2 T^{-2} A^{-2}]$ |
| C. Magnetic permeability | III. $[M L^0 T^{-2} A^{-1}]$ |
| D. Self inductance | IV. $[M L^2 T^{-2} A^{-1}]$ |
Choose the correct answer from the options given below: