Excess Force to Lift a Ring – Rankers Physics

Surface Tension and Viscosity: Practice Problem & Solution

If surface tension of a liquid is \(T\), then excess force required to lift a ring of radius \(R\) from surface of this liquid is:
\(T \times 2\pi R\)
\(T \times \pi R\)
\(T \times 4\pi R\)
\(T \times 8\pi R\)

Solution Explained:

To solve this problem, we apply the core principles of Surface Tension and Viscosity. Understanding the underlying formula is key to arriving at the correct answer below:

A ring has two free surfaces in contact with the liquid (inner and outer boundaries). Therefore, the total contact length is \(L = 2(2\pi R) = 4\pi R\), and the excess force required is \(F = T \times 4\pi R\).

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