The surface tension of a soap solution is $2 * {10^{ - 2}}N/m$ . To blow a bubble of radius 1 cm, the work done is
(A) $4\pi * {10^{ - 6}}J$
(B) $8\pi * {10^{ - 6}}J$
(C) $12\pi * {10^{ - 6}}J$
(D) $16\pi * {10^{ - 6}}J$

Hint: We are aware that every liquid has a propensity to condense to the smallest surface that is feasible given the available resources. Surface tension of the liquid is the name given to this characteristic of liquids. This kind of quality may be seen in the soap solution in the question.

Formula used:
$W=T\times \Delta A$
W is the work done, T is the surface tension and $\Delta A$is the change in area

Complete answer:
Start with the formula of the amount of work done in the case of soap solution. We know that, Work done is equal to tension in surface energy and surface energy is tension multiply to change in area as follows:
$W = T \times \Delta A$
Where,
W is work done
T is surface tension and
$\Delta A$is change in area.

Now, in given case of soap bubble, work done will be;
$W = 4\pi {r^2} \times 2 \times T = 8\pi {r^2}T$

From the question, we know that;
$T = 2 \times {10^{ - 2}}N/m$
And $r = 1cm$

Putting the value in above equation of work done, we get;
$W = 8\pi \times {\left( {{{10}^{ - 2}}} \right)^2} \times 2 \times {10^{ - 2}}$
$W = 16\pi \times {10^{ - 6}}J$

Hence the correct answer is Option D.

Note: One needs to be familiar with surface tension ideas and applications, as well as the effort needed to raise or lower any object's surface tension, in order to answer the provided problem. Additionally, it's important to recall the mathematical relationship between work and surface tension. Along with the participation of the surface energy, the surface tension is essential to the development and analysis of the soap bubbles.