Question

(I) If the force \(F\) needed to move the wire in Fig. 10–34 is \(3.4 \times {10^{ - 3}}\;{\rm{N}}\), calculate the surface tension \(\gamma \) of the enclosed fluid. Assume \(l = 0.070\;m\).

Short answer

The surface tension of the enclosed fluid is \(0.024\;{\rm{N/m}}\).

Step-by-step solution

Concept

Assuming a U-shaped apparatus is enclosing a thin film of liquid. The free-body diagram of the apparatus is given below:

Here, \(F\) is the surface tension force applied on the wire, \(l\) is the length of the wire and \(\gamma \) is the surface tension of the enclosed fluid.

The liquid enclosed in the apparatus consists of two surfaces, a bottom surface and a top surface. So, the total length of the surface under tension is twice the initial length.

The expression for the surface tension of the enclosed fluid is given by,

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

Given data

The force needed to move the wire is, \(F = 3.4 \times {10^{ - 3}}\;{\rm{N}}\).

The length of the wire is, \(l = 0.070\;{\rm{m}}\).

Calculation

The expression for the surface tension of the enclosed fluid is given by,

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

Substituting the values in the above expression,

\(\begin{array}{l}\gamma = \frac{{3.4 \times {{10}^{ - 3}}\;{\rm{N}}}}{{2 \times 0.070\;{\rm{m}}}}\\\gamma = 0.024\;{\rm{N/m}}\end{array}\)

Therefore, the surface tension of the enclosed fluid is \(0.024\;{\rm{N/m}}\).