Answer:
P₂ = 4098.96 Pa
Step-by-step explanation:
For this exercise let's use Bernoulli's equation
Let's use the subscript 1 for the point of the liquid surface and the subscript 2 for the ends (point A)
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
the velocity at the end of the tank
v₂ = w r₂
the velocity at the surface of the liquid is
v₁ - w r₁
where r₂ = 1.5 m and r₁ = 1 m
the tank pressure is P₁ = P₀ = 0.1 10³ Pa
P₂ = P₁ + ½ ρ [w² (r₁² - r₂²)] + ρ g (y₁ -y₂)
We must remember that the pressure measurements the distances are measured from the lowest part to the surface that has zero height
let's reduce the magnitudes to the SI system
w = 3 rev / min (2π rad / 1rev) (1 min / 60 s) = 0.314159 rad / s
let's calculate
P₂ = 0.1 10³ + ½ 800 0.314159² |(1² -1.5²)| + 800 9.8 |(1-1.5)|
P₂ = 0.1 103 +78.96 + 3920
P₂ = 4098.96 Pa