How much work will be done in making a soap bubble of diameter \[2.0\;{\rm{cm}}\].
A. \[8.54 \times {10^{ - 5}}\]
B. \[7.54 \times {10^{ - 5}}\]
C. \[9.54 \times {10^{ - 5}}\]
D. \[10.54 \times {10^{ - 5}}\]

Hint: The above problem can be resolved by applying the concept and mathematical formula of the work done to make the soap bubble. The mathematical relation is given by taking the product of surface tension and area of the bubble. The standard value for the surface tension of the soap bubble can be taken to make the solution more reliable.

Complete step by step answer:
Given:
The diameter of the soap bubble is, \[d = 2.0\;{\rm{cm}}\].
The mathematical formula for the work done to make the soap bubble is given as,
\[W = \sigma \times A\]
Here, \[\sigma \] is the surface tension of the soap bubble and its standard value is \[3 \times {10^{ - 2}}\;{\rm{N/m}}\].And A is the area of the soap bubble and its value is,
\[A = 4\pi {r^2}\]
Here, r denoted the radius of the soap bubble and its value is, \[r = d/2\].
Solve by substituting the values as,
 \[\begin{array}{l}
W = \sigma \times A\\
W = 3 \times {10^{ - 2}}\;{\rm{N/m}} \times \left( {4\pi {r^2}} \right)\\
W = 3 \times {10^{ - 2}}\;{\rm{N/m}} \times \left( {4\pi {{\left( {\dfrac{d}{2}} \right)}^2}} \right)\\
W = 3 \times {10^{ - 2}}\;{\rm{N/m}} \times \left( {4\pi {{\left( {\dfrac{{20\;{\rm{cm}} \times \dfrac{{{{10}^{ - 2}}\;{\rm{m}}}}{{1\;{\rm{cm}}}}}}{2}} \right)}^2}} \right)\\
W = 7.54 \times {10^{ - 5}}\;{\rm{J}}
\end{array}\]
Therefore, the required work to make the soap bubble is \[7.54 \times {10^{ - 5}}\;{\rm{J}}\].And option B is correct.

Note:In order to resolve the given problem, one must be aware of the concepts and applications of the surface tension, along with the work required to increase or decrease the surface tension of any object. Moreover, the mathematical relation of work and surface tension is to be remembered. The surface tension plays a vital role in the formation and the analysis of the soap bubbles, along with the involvement of the surface energy.