Question

A circular, thin metal ring of radius 0.022 m and mass 0.014 kg is submerged in...

A circular, thin metal ring of radius 0.022 m and mass 0.014 kg is submerged in a fluid. The ring is constrained to be held in a horizontal position when it is pulled upward. When the force lifting the ring reaches 0.25 N, the ring pulls free of the fluid. What is the coefficient of surface tension of the fluid in SI units of force per length? Calculate the answer to three decimals. (Hint: What is the length of metal in contact with the fluid?)

the answer is 0.408, but I want the steps

The surface tension coef.() is simply the tension force per metre acting perpendicular on a surface ..

(N/m)

In this case, the tension force is acting perpendicular downwards on the ring at the moment it pulls free .. there are 2 meniscus surfaces meeting the ring vertically (against the inner and outer circumferences)

As the ring pulls free the upward and downward forces acting are in balance ..
Upward force F↑= 0.25N
Downward forces F↓ = weight of ring (mg) + s.t force (L) ......L=2R

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