A right–circular cylinder, open at the top, is filled with the liquid of relative density 1.2. It is rotated about its vertical axis at such a speed that half of the liquid spills out. The pressure at the center of the bottom will be
- AZeroCorrect
- BOne-fourth its value when cylinder is full
- COne-half its value when cylinder is full
- DCannot be predicted from the given data
Solution
Concept
Forced vortex in an open cylinder
Formula
Volume of paraboloid of revolution = ½ × volume of the circumscribing cylinder
Step-by-step solution
1. When rotated, the free surface becomes a paraboloid. 2. The empty space above the paraboloid is a paraboloid of volume ½ × (πR²h). 3. Half the liquid spilled, so the remaining liquid is exactly the volume below the paraboloid: its vertex touches the bottom. 4. The free surface at the centre is at the bottom, so the depth of liquid over the centre is zero. 5. Pressure at the centre of the bottom = ρ g × 0 = 0 (gauge).
Explanation
For a cylinder rotating at high speed, the paraboloid cuts the cylinder and the water left is half. The vertex of the free surface lies at the base, so there is no liquid above the central point and the gauge pressure there is zero.
Shortcut
Half the liquid spills = vertex touches the bottom = zero pressure at the centre.
Common mistake
Trying to calculate with ρ = 1.2 × 1000 kg/m³ and some depth.
Correct answer: A) Zero