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Question #42095

Difficulty: 8/10 | 🟢 Free

The Question

A hollow rigid sphere of inner radius RR is completely filled with an incompressible liquid of density ρ\rho. The sphere rotates with a constant angular velocity ω\omega about a vertical diameter. A small hole at the highest point of the sphere keeps the liquid pressure there equal to the atmospheric pressure P0P_0. Assuming the acceleration due to gravity is gg, what is the net upward force exerted by the liquid on the upper half of the sphere?

Options

  • A.

    πR2P0+13πρgR3+π4ρω2R4\pi R^2 P_0 + \frac{1}{3} \pi \rho g R^3 + \frac{\pi}{4} \rho \omega^2 R^4

  • B.

    πR2P0+23πρgR3+π4ρω2R4\pi R^2 P_0 + \frac{2}{3} \pi \rho g R^3 + \frac{\pi}{4} \rho \omega^2 R^4

  • C.

    πR2P0+13πρgR3+π2ρω2R4\pi R^2 P_0 + \frac{1}{3} \pi \rho g R^3 + \frac{\pi}{2} \rho \omega^2 R^4

  • D.

    πR2P0+23πρgR3+π2ρω2R4\pi R^2 P_0 + \frac{2}{3} \pi \rho g R^3 + \frac{\pi}{2} \rho \omega^2 R^4

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Centripetal Acceleration Vector|Accelerated Frame Hydrostatics|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the application of hydrostatics in an accelerated reference frame, specifically targeting the trap of 'Pressure Gradient Fluctuation'.

Students often incorrectly assume that the pressure at the equatorial plane is uniform and equal to its maximum value at the edge (P0+ρgR+12ρω2R2P_0 + \rho g R + \frac{1}{2}\rho \omega^2 R^2), leading them to multiply this by the total area πR2\pi R^2.

This ignores the radial pressure gradient ∂P/∂r=ρω2r\partial P/\partial r = \rho \omega^2 r, which requires integration over the circular cross-section.

Furthermore, students frequently miscalculate the net vertical force from the static fluid by confusing the Archimedes buoyant force with the actual integrated pressure force, incorrectly using the hemisphere's weight 23πR3ρg\frac{2}{3}\pi R^3 \rho g as the sole gravitational contribution instead of evaluating the boundary equilibrium Fup=Feq−WF_{up} = F_{eq} - W.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterMotion in a Plane - Class 11 Part 1
ConceptCentripetal Acceleration Vector
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree