(II) The surface tension of a liquid can be determined by measuring the force \(F\) needed to just lift a circular platinum ring of radius \(r\) from the surface of the liquid. (a) Find a formula for \(\gamma \) in terms of \(F\) and \(r\). (b) At \(30^\circ C\), if \(F = 6.20 \times {10^{ - 3}}\;{\rm{N}}\) and \(r = 2.9\;{\rm{cm}}\), calculate \(\gamma \) for the tested liquid.

Surface tension is defined as the amount of force that acts on the unit surface and it resists all the external force acting on the surface.

The expression for the surface tension is given by,

\(\gamma = \frac{F}{{2L}}\)

Here, \(\gamma \) is the surface tension, \(F\) is the force applied, and \(L\) is the total length (circumference) of the surface.

The force applied on the surface is, \(F = 6.20 \times {10^{ - 3}}\;{\rm{N}}\).

The radius of the circular ring is, \(r = 2.9\;{\rm{cm}} \times \frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}} = 0.029\;{\rm{m}}\).

Part (a)

The expression for the surface tension acting on the circular platinum ring is given by,

\(\begin{array}{l}\gamma = \frac{F}{{2L}}\\\gamma = \frac{F}{{2\left( {2\pi r} \right)}}\\\gamma = \frac{F}{{4\pi r}}\end{array}\)

Here, \(r\) is the radius of the circular platinum ring.

Thus, the above formula is used to calculate surface tension \(\gamma \) in terms of \(F\) and \(r\).

Part (b)

The expression for the surface tension for the tested liquid is given by,

\(\gamma = \frac{F}{{4\pi r}}\)

Substituting the values in the above expression,

\(\begin{array}{l}\gamma = \frac{{6.20 \times {{10}^{ - 3}}\;{\rm{N}}}}{{4\pi \times 0.029\;{\rm{m}}}}\\\gamma = 0.017\;{\rm{N/m}}\end{array}\)

Thus, the surface tension for the tested liquid is \(0.017\;{\rm{N/m}}\).