Question

A film of water is formed between two straight parallel wires each 10 cm long and at a separation of $0.5cm$ . Calculate the work required to increase the distance between them by 1 mm
Surface tension of water $= 72 \times {10^{ - 30}}{N \mathord{\left/ {\vphantom {N m}} \right.} m}$ .

Answer 1

Hint : We need to find out how the change in area affects the change in surface energy. Work done will be given by the change in the surface energy.

Formula Used: The following formulas are used to solve this question.
$\Rightarrow W = \Delta S \times T$ where $W$ is the work done, $\Delta S$ is the increase in surface area and $T$ is surface tension of water.
$\Rightarrow S.E. = S \times A$ where $S.E.$ is the surface energy, $S$ is surface tension and $A$ is area.

Complete step by step answer
Work done by a body is the product of the force applied in the direction of displacement, on it and the resulting displacement due to the force.
$\Rightarrow W = F \cdot d$ where $F$ is the force, $d$ is displacement and $W$ is the work done.
In other words, it is the energy transferred when an object is moved over a distance by an external force applied in the direction of the displacement.
If the force is being exerted at an angle $\theta$ to the displacement, the work done is $W = F \cdot d\cos \theta$
Work done on a body is accomplished not only by a displacement of the body as a whole from one place to another but also, for example, by compressing a gas, by rotating a shaft, and even by causing invisible motions of the particles within a body by an external magnetic force.
Work done in this case is the same as the change in surface energy.
Now, surface energy is the product of surface tension and area.
Surface energy $S.E. = S \times A$ where $S$ is surface tension and $A$ is area.
Since the value of surface tension cannot change, change in surface energy must take place due to change in area.
$\therefore \Delta S.E = S \cdot \Delta A$
Given in the question, the increase in width is $0.1cm$ . The length of the parallel wires is given and is equal to 10cm.
Original area $\Delta A = l \times \Delta w$ where $l$ is the length and $w$ is the width.
$\Rightarrow$ $\Delta A = 2\left[ {10\left. { \times \left( {0.1} \right)} \right]} \right.cm$
$\Rightarrow \Delta A = 2cm = 0.02m$
Change in surface energy $\Delta S.E = S \cdot \Delta A$
Assigning the values given in the question, $S = 72 \times {10^{ - 30}}{N \mathord{\left/ {\vphantom {N m}} \right.} m}$ and $\Delta A = 0.02m$ ,
$\Rightarrow \Delta S.E. = 72 \times {10^{ - 3}} \times 0.02J$
$\Rightarrow \Delta S.E. = 1.44 \times {10^{ - 5}}J$ .

Note
Surface tension is the physical force acting on the surface of a liquid such that it pulls in its neighbouring molecules from all directions, arising from the cohesive nature of liquid molecules. Surface energy results in disrupted intermolecular bonds when a film is created.