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

AI Thread Summary
The discussion revolves around calculating the work done to double the radius of a soap bubble, given its surface tension. The initial surface energy is calculated as Ui = 4πr²S, while the final surface energy after doubling the radius is Uf = 16πr²S. The change in surface energy, which represents the work done, is determined to be Uf - Ui = 12πr²S. However, the correct answer is identified as 24πr²S, as the surface tension acts on both sides of the bubble, effectively doubling the surface tension to 2S. The misunderstanding was clarified, leading to a resolution of the problem.
vissh
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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 ^.^
 
Physics news on Phys.org
For a bubble surface tension is on its two sides i.e. 2S :smile:
 
lol . I missed that xD Thanks abdul :) Got it ^.^
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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