Surface Energy Calculation: 8-24(πr^2S)

[vissh]

hiii :)

Homework Statement

(Q)Air is pushed into a soap bubble of radius r to double its radius.If the surface tension of the soap solution is S, the work done in the process is
<a>8(pie)(r)²(S) <b>12(pie)(r)²(S) <c>16(pie)(r)²(S)
<d>24(pie)(r)²(S)

Homework Equations

Surface energy of a liquid = (S)(A)
where S is surface tension and A is the area of the surface of liquid.

The Attempt at a Solution

Taking the bubble as system.
Isn't it like this:-
>Work done in process i.e. work done by external agent = change in Potential energy = change in surface energy(U_f - U_i)
> U_i = 4(pie)(r)²(S)
> U_f = 4(pie)(2r)²(S) = 16(pie)(r)²(S)]
> U_f - U_i = 12(pie)(r)²(S) = Work done we require
But the answer is <d>
Can any1 help me where i am getting wrong :)
Thanks for reading ^.^

 

[zorro]

For a bubble surface tension is on its two sides i.e. 2S

 

[vissh]

lol . I missed that xD Thanks abdul :) Got it ^.^