Project ForgeA large tank with a nozzle attached contains three immiscible, inviscid fluids of densities ρ1, ρ2, ρ3 (top to bottom) in layers of depth h1, h2, h3, with the nozzle at the bottom discharging fluid ρ3 to atmosphere. Assuming that the changes in h1, h2 and h3 are negligible, the instantaneous discharge velocity is
- √(2gh3(1 + (ρ1/ρ3)(h1/h3) + (ρ2/ρ3)(h2/h3)))
- B√(2g(h1+h2+h3))
- C√(2g(ρ1h1+ρ2h2+ρ3h3)/(ρ1+ρ2+ρ3))
- D√(2g(ρ1h2h3+ρ2h3h1+ρ3h1h2)/(ρ1h1+ρ2h2+ρ3h3))
Explanation
The static pressure at the nozzle (base of layer 3) is the sum of each layer's weight per unit area: P=ρ1gh1+ρ2gh2+ρ3gh3. Since the fluid actually issuing from the nozzle has density ρ3, apply Torricelli's law with THAT density: V=√(2P/ρ3)=√(2g(ρ1h1+ρ2h2+ρ3h3)/ρ3). Factoring out h3 inside the root: =√(2gh3(1+(ρ1h1)/(ρ3h3)+(ρ2h2)/(ρ3h3))), matching option (A) exactly.
Source: GATE ME 2012 (Booklet Code A) (2012)
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