15-10-09 Surface tension

Suppose that a bubble has the shape of a long cylinder, rather than that of a sphere. Determine an expression for the difference between the inside and outside pressures; express your answer in terms of the surface tension γ and the radius R of the cylinder

PS: Solving this question explains a lot of very basic things about surface tension :) .. Do not ignore ;)

Source: Internet...

18 Answers

3
iitimcomin ·

@eure that i know ......ur looking only at the force coused by T in that interface .... that 2r is part of the actual surface of bubble .. there is no proj here .....

62
Lokesh Verma ·

yup... That's correct..

24
eureka123 ·

ohh...ya i took only one side of cylinder..
so I just need to multiply my result with 2 naa ??

62
Lokesh Verma ·

The answer will be like this..

You have all made some slight mistakes..

Eureka has taken only one film...

WHile IITimcomin has made a mistake of taking force due to pressure as P(2piRL) The logic of which is nto clear to me..

According to me the method should be

Pi-Po(2RL) = T(2L+2L+4r+4R)

So Pi-Po = T (2L+4R) / LR = 2T/R + 4T/L

since L is very large, it is same as 2T/R

(Here I have assumed pure cyllindrical shape which may not be ture.. but it wont matter because the lenght is very large in comparison to the radius...)

24
eureka123 ·

Anyways answer remians same..becoz answer didnt depend on that term (L→∞)[3]

1
xYz ·

@eureka surface tenion is like kind of internal force in between the two halves.......so it go to act over the perimeter of the rectangle ......:)

24
eureka123 ·

ok realising my msitake now..

3
iitimcomin ·

xyz he said ur ans is rong cuz u didnt take L>>R... else it shud be right!!!!

24
eureka123 ·

@xyz
Tension force wont act on projection.....it will act on pure bubble surface

so pure bubble surface will have perimeter...Ï€R+Ï€R+2L

62
Lokesh Verma ·

are you sure karna... (It is given that L is very large... )

So the answer will be independent of L

Try again.. (not that this needs to be necessariliy true)

1
xYz ·

@eureka how are u taking the perimeter to be 2pir+2l.........
the cross section will be a rectangle of dimensions 2r and l
corect me if i am wrong

24
eureka123 ·

@ iitimcomin
Tension force wont act on projection.....it will act on pure bubble surface

@xyz
ur terms are diff

1
xYz ·

even i have done the same thing in first post...but nishant sir has told that its wrong

3
iitimcomin ·

eure u made a mistake...

it should be

T(2(2R+L))..........its the perimeter of the recangle u get as crossection.... think abt it.....

so taking R<<L u get T/R..........[FOR LIQUID]........

so for soaP... u get 2T/R!!!

#6 is another way f doin it u get same ans...[1]

24
eureka123 ·

I considered it to be constitued of liquid not soap

24
eureka123 ·

Cutting the cylinder through middle

Pout(2RL)+T(Ï€R+Ï€R+2L)=P(2RL)
=>(P-Pout)=T(2Ï€R+2L)2RL
=>(P-Pout)=T(Ï€R+L)RL
=>(P-Pout)=T[Ï€L+1R]

L→∞
=>(P-Pout)=TR

Correct soln

Pout(2RL)+T(2R+2R+2L)=P(2RL)
=>(P-Pout)=T(4R+2L)2RL
=>(P-Pout)=T(2R+L)RL
=>(P-Pout)=T[2L+1R]

L→∞
=>(P-Pout)=TR

also edit :
Put γ in palce of T

3
msp ·

me getting Pex=2Y/R

Since the L>>>R

we have Y.(πLΔr+πΔLR)=(ΔP).(2ΠLrΔr+πΔLR)

which yields us ΔP=2Y/r

3
iitimcomin ·

let there be excess pressure .P ... inside .....

therefore .. we have ..
P(2piRL)dr = 2T(2piLdr + 2piRdr)

P = 2T(L+R)/(RL)

if L>>R

we have P = 2T/R

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