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This question has intriqued me because two methods are generally employed to derive the shape of the surface(paraboloid) .

Both methods give equal numerical answer but differ in which of them P or Q (in figure) has greater pressure

(1) Firstly as done in the book one applies Bernoulli and finds lesser pressure at Q due to it having more velocity.

(2) One can calculate net force (gravity and centrifugal) at a point x distance away from centre and make the surface's tangent perpendicular to it , giving us a solvable differential equation.

The second method predicts more pressure at Q due to it being towards the direction of centrifugal force.

Also if one considers a point R above Q at the surface what will be its pressure.

Logic says , it is equal to pressure at P because both are at surface , but then pressure at Q is greater than at R ( it is below R) so doesnt that mean pressure at Q is greater than at P.

Please explain this situation concluding about exact pressures at P , Q and R and their corresponding reasons.

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closed as off-topic by user191954, Kyle Kanos, Jon Custer, Aaron Stevens, ZeroTheHero Apr 4 at 4:40

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  • $\begingroup$ Do you read study material by RESONANCE? $\endgroup$ – user213933 Mar 23 at 13:10
  • $\begingroup$ @Shreyansh No this is Aakash Study Material $\endgroup$ – user224768 Mar 23 at 13:53
  • $\begingroup$ Are you in class 11? $\endgroup$ – user213933 Mar 23 at 14:20
  • $\begingroup$ No I am about to finish Class XII $\endgroup$ – user224768 Mar 24 at 1:34
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Bernoulli works for streamlines which in this case are circles of constant radius and constant vertical height.

This means that a line from $P$ to $Q$ is not a streamline and so the analysis in the book is flawed.

To apply Bernoulli in a useful way you must move to the rotating frame in which the fluid is at rest.
In such a rotating frame any two points are connected by a streamline and so you can use the line joining $P$ and $Q$ as a streamline and with the introduction of a fictitious force can use Bernoulli and show that the pressure at $Q$ is greater than the pressure at $P$.

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