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Surface energy

  • Thread starter vissh
  • Start date
  • #1
82
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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(Uf - Ui)
> Ui = 4(pie)(r)2(S)
> Uf = 4(pie)(2r)2(S) = 16(pie)(r)2(S)]
> Uf - Ui = 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 ^.^
 

Answers and Replies

  • #2
1,384
0
For a bubble surface tension is on its two sides i.e. 2S :smile:
 
  • #3
82
0
lol . I missed that xD Thanks abdul :) Got it ^.^
 

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