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find ū at hole.
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ρ = Density of the liquid
A = Area of cross section ( hole )
P = 1 atm
P₀ = Liquid pressure
By equation of continuity,
A₁ v₁ = A₂ v₂
v₂ = A₁ v₁ / A₂
As A₂ >> A₁
v₂ ( ū ₁ ) = 0
Applying Bernoulli's Equation,
P₀ + ½ ρ v₁ ² + ρ g( h₂ - h₁ ) + ( P - P₀)
If we take h₂ - h₁ = h, then
½ ρ v₁ ² = ρgh + ( P₀- P )
v₁ ( ū ) = √ 2gh + [ 2 ( P₀ - P) / ρ ]
=> v₁ ( ū ) = √ 2gh + [ 2 P₀ / ρ ] { ∴ P = 1 atm }
If the container is open, then
v₁ ( ū ) = √ 2gP₀ / ρ
A = Area of cross section ( hole )
P = 1 atm
P₀ = Liquid pressure
By equation of continuity,
A₁ v₁ = A₂ v₂
v₂ = A₁ v₁ / A₂
As A₂ >> A₁
v₂ ( ū ₁ ) = 0
Applying Bernoulli's Equation,
P₀ + ½ ρ v₁ ² + ρ g( h₂ - h₁ ) + ( P - P₀)
If we take h₂ - h₁ = h, then
½ ρ v₁ ² = ρgh + ( P₀- P )
v₁ ( ū ) = √ 2gh + [ 2 ( P₀ - P) / ρ ]
=> v₁ ( ū ) = √ 2gh + [ 2 P₀ / ρ ] { ∴ P = 1 atm }
If the container is open, then
v₁ ( ū ) = √ 2gP₀ / ρ
Is it correct?
yep
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