Question

The amount of work done in blowing a soup bubble such that its diameter increases from $d\,to\,D$ is: (surface tension of solution is T)
(A) $4\pi ({D^2} - {d^2})T$
(B) $8\pi ({D^2} - {d^2})T$\[\]
(C) $\pi ({D^2} - {d^2})$
(D) $2\pi ({D^2} - {d^2})T$

Answer 1

Hint: First start to find out what will be the relation between the amount of work done in blowing a soup bubble, its diameter and the surface tension then try to put all the information provided in the question in that relation. Also the D is greater than d as given in the question so put its value accordingly.

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 soap 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;
$W = 8\pi T(r_2^2 - r_2^2)$ (equation 1)

From the question, we know that;
${r_1} = \dfrac{d}{2}\,and\,{r_2} = \dfrac{D}{2}$

Putting the value of \[{r_{1\,}}\,and\,{r_2}\] in equation 1, we get;
\[W = 8\pi T\{ (\dfrac{{{D^2}}}{4}) - (\dfrac{{{d^2}}}{4})\} \]

After solving, we get;
\[W = 2\pi T({D^2} - {d^2})\]

Hence the correct answer is Option D

Note: The thing to be kept in mind while talking about the surface tension. Fluids have a feature called surface tension that causes the fluids to tend to minimize or shrink their surface area. The force acting along the length of a fluid's surface is known as surface tension.