What is the mass of the water in the stirrer supported by surface tension?

[Galactium]

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

 

[TSny]

Your work looks good. What numerical answer did you get for m?

 

[Galactium]

Your work looks good. What numerical answer did you get for m?

Nevermind, I got the right answer just a minor mistake. Sorry about that. Thanks for the help!