Question

Air is pushed into a soap bubble of radius r to double its radius. If the surface tension of the soap solution in S, the work done in the process is
(a) 8 π r2 S
(b) 12 π r2 S
(c) 16 π r2 S
(d) 24 π r2 S

Answer 1

The value of the work done in the process is 24πr^2 S.

Explanation:

No. of surfaces of a soap bubble = 2

Increase in surface area = 4π(2r)^2 − 4π(r)^2

                                         = 12πr^2

Total increase in surface area = 2×12πr2

                                         =24πr^2

Work done = change in surface energy                  

                                =S × 24πr^2

                                = 24πr^2 S

Thus the value of the work done in the process is 24πr^2 S.

Also learn more

Please solve Volume Surface Area Questions 17,18,19,20 .

https://brainly.in/question/6979878

Answer 2

The work done in the process is $24 \pi r^{2} S$ with the surface tension of the soap solution in S.

Explanation:

A soap bubble's no. of surfaces = 2

Increase in the surface area = $4 \pi(2 r)^{2}-4 \pi(r)^{2}=12 \pi r^{2}$

Total rise in surface area = $=2 \times 12 \pi r^{2}=24 \pi r^{2}$

It is to be understood that the work done is equal to the change in surface energy.

So, $S \times 24 \pi r^{2}=24 \pi r^{2} S$