The pressure in a cylindrical jet of water with a diameter of 4mm exceeds the surrounding atmospheric pressure due to surface tension. The correct formula for calculating this pressure difference is ΔP = σ/r, where σ is the surface tension (0.0718 N/m) and r is the radius (2mm). The initial confusion regarding the formula ΔP = 2σ/r was clarified, confirming that the correct application of the law is indeed ΔP = σ/r.
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On what basis do you feel it should be ΔP= 2σ/r?Amr719 said:Homework Statement
By How much does the pressure in a cylinderical jet of water 4mm in diameter exceed the pressure of the surrounding atmosphere if the surface tension of water is 0.0718 N/M?
Homework Equations
The Attempt at a Solution
ΔP= σ/r=0.0718/2*10^-3 , This is the answer I have but I don't understand from where did he get this law . Isn't it supposed to be ΔP= 2σ/r instead ?[/B]
Sorry I was wrong , I understand it nowChestermiller said:On what basis do you feel it should be ΔP= 2σ/r?
Chet