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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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