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A closed cylinder having a radius R and height H is filled with oil of density ρ. If the cylinder is rotated about its axis at an angular velocity of ω, the thrust at the bottom of the cylinder is
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- πR2ρ.gH
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πR2 ρ ω2 R2 4 - πR2(ρ ω2R2 + ρ gH)
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πR2 
ρ ω2R2 + ρ gH 
4
Correct Option: D
We know that
| = | = | = δω2r | ||||
| δr | r | r |
[∵ v = ω × r]
| p | δp = | | r | ρω2r dr | 0 | 0 |
| p = | ω2r2 | 2 |
Area of circular ring = 2πrdr
Force on elementry ring
= Intensity of pressure × Area of ring
| = | ω2r22πr dr | 2 |
∴ Total force on the top of the cylinder

| = | | R | ρω2r22π r dr | 0 | 2 |
| = | ρω2r22π | | R | r3 dr | 2 | 0 |
| . ω22π | = | ω2 × πR4 | ||||
| 2 | 4 | 4 |
Thrust at the bottom of the cylinder
= Weight of water in cylinder + Total force on the top of cylinder
| = ρg × πR2 × H + | ω2 × πR4 | 4 |
| = πR2 | ![]() | + ρgH | ![]() | ||
| 4 |

