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Galactium
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Homework Statement
A hollow stirrer that has a diameter of 2.0 mm is inserted into a cup of water at 20∘C. The surface tension of water at that temperature is γ(surface tension) = 7.28×10−2 N/m.
What is the mass of the water supported by the surface tension in the stirrer?
Given:
d=2 mm
p(rho): 1000 kg/m^3
y(surface tension): 7.28E-2 N/m
Find: Mass
Conversions:
d=.002 m r=.001
Problem: I tried this equation, but I keep on getting the wrong answer, I am not sure if I am using the correct formula.
Homework Equations
Surface tension: deltaP=2y/r
Pressure: deltaP=F/A
The Attempt at a Solution
detlaP=2y/r
deltaP=F/A
F=mg
A=(pi)r^2
Plug them in:
mg/(pi)r^2=2y/r
Change the equation to equal mass (m):
m=2y(pi)r/g
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