Question

The work done in blowing a soap bubble of 10 cm radius. And the surface tension of soap solution is $0.03N/m$ .
A. $37.68 \times {10^{ - 4}}J$
B. $75.36 \times {10^{ - 4}}J$
C. $126.82 \times {10^{ - 4}}J$
D. $75.36 \times {10^{ - 3}}J$

Answer 1

Hint: By using the concept and mathematical formula used to create the soap bubble, the aforementioned issue can be solved. Taking the surface tension and bubble area as a product yields the mathematical relationship. To increase the solution's dependability, the standard value for soap bubble surface tension might be used.

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 blowing a soup bubble. 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 the change in area.

Now, in given case of soap bubble, work done will be;
$W = 4\pi {r^2} \times 2 \times T = 8\pi {r^2}T$ (equation 1)

From the question, we know;
$T = 0.03N/m$
$r = 10cm$

Putting these value in equation 1, we get;
$W = 8\pi \times {\left( {10 \times {{10}^{ - 2}}} \right)^2} \times 0.03$

After solving, we get;
$W = 75.36 \times {10^{ - 4}}J$
Hence the correct answer is Option B.

Note: We should be aware that the force that the molecules under the surface of the liquid exert on the molecules on the liquid's surface is what causes the property of surface tension to form. The liquid's molecules constantly have a propensity to pull the liquid's surface molecules into the bulk of the liquid. As a result, the molecules have a tendency to contract and occupy the smallest surface area. This surface tension force, which functions as a sort of cohesive force, is only felt while facing inward.