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Surface tension homework

  1. Aug 1, 2009 #1
    1. The problem statement, all variables and given/known data
    A glass tube is vertically suspended using a spring balance.The thickness of the glass tube is 0.4mm.Now it is lowered towards a liquid.When the lower surface of the tube touches the liquid surface,a change can be seen in the spring balance reading.
    Now the tube is immersed in the liquid until the reading of the spring balance returns to its original reading.
    At this stage the height of the tube immersed in the liquid is 3.67cm.If the density of the liquid is 1000kgm-3,calculate the surface tension of the liquid.
    [ans:7.34*10-2]



    2. Relevant equations

    T(surface tension)=F/l

    3. The attempt at a solution
    l or the length the surface tension acts on = 2*pi*r + 2*pi(r+d)
    where r=internal radius of the tube
    r+d=external radius of the tube and d=0.4mm

    F=T*l
    F=2*pi*T(2r+d)


    When the tube just touches the liquid the scale reading increases(to S) due to surface tension forces acting down on the tube.
    When the tube is immersed in the fluid,upthrust probably acts on the tube restoring the original scale reading(R) ?
    But I'm not sure how this helps me find T
     
  2. jcsd
  3. Aug 1, 2009 #2

    rl.bhat

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    Is it not necessary to know the radius of the tube r to find the surface tension?
     
  4. Aug 1, 2009 #3
    I do not know,but the value for r was not given in the question.
     
  5. Aug 1, 2009 #4
    Are you sure you presented the question correctly?I suspect that it is a glass rod of radius 0.4mm.If so when balance is restored the upthrust( pi*r^2*h*density*g) is balanced by the force of surface tension(2*pi*r*surface tension) which pulls down on the outer circumference of the rod.By the way the liquid is water.
     
  6. Aug 1, 2009 #5
    Yes.That is exactly how our teacher gave us the question.
    The question:
    A glass tube is vertically suspended using a spring balance.The thickness of the glass tube is 0.4mm.Now it is lowered towards a liquid.When the lower surface of the tube touches the liquid surface,a change can be seen in the spring balance reading.
    Now the tube is immersed in the liquid until the reading of the spring balance returns to its original reading.
    At this stage the height of the tube immersed in the liquid is 3.67cm.If the density of the liquid is 1000kgm-3,calculate the surface tension of the liquid.

    Ah!yes. the upthrust = force of surface tension
    Then for a tube would the upthrust be= pi*(r+d)^2*h*density*g ?
     
  7. Aug 1, 2009 #6
    I tried working with Upthrust(U)=pi*(r+d)^2*h*density*g and surface tension force(F)=2*pi*T(2r+d),
    and I end up with (r2+2rd+d2)/2(2r+d) ,but I don't know how I can solve this without knowing r?

    Then I tried using U=[pi*(r+d)2 - pi*r2]*h*density*g
    U=pi[(r+d2)-r2]h*density*g
    U=pi*(2r+d)*d*h*density*g

    When I equate this to the surface tension force ,I get the answer,but to find upthrust don't we have to take the whole volume of the part of the object immersed?
     
  8. Aug 1, 2009 #7
    I just assumed it was a solid,non hollow,rod and used r not r+d (my assumption being based on the apparent fact that information seemed to be missing from the question).You do need to take into account the whole volume submerged and the upthrust is pi*r^2*h* density*g( h= length of rod submerged)
     
  9. Aug 1, 2009 #8
    Yes,we do get the answer for a tube when we use
    ( pi*r^2*h*density*g) = (2*pi*r*surface tension) assuming 0.4mm is the diameter of the tube.

    But I don't know.I just feel,my sir would have told us if it were a glass rod and not a tube.
    (I wonder if it can be a mistake,we've used the word 'tube' at least 3 times in the question)
    I'll try to check with the others and see if it was a mistake afterall.
    Anyway thanks for the help!
     
  10. Aug 1, 2009 #9

    rl.bhat

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    The volume of the displaced liquid can be written as
    V =π[( r + d)^2 - r^2]*h
    = π[(2r+d)*d*h
    So π[(2r+d)*d*h*density*g = 2*π*[2r+d]*surface tension.
    Or
    T= d*g*h*density/2
    Now find T.
     
  11. Aug 1, 2009 #10
    I get the answer for T,but I don't understand how the volume of liquid displaced can be taken as V =π[( r + d)^2 - r^2]*h.
    Why have we ignored the volume displaced by the whole tube?Usually to find the upthrust acting on an object don't we include the volume of the cavity as well in our calculations?
     
  12. Aug 1, 2009 #11

    rl.bhat

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    The upthrust is equal to the weight of the displaced liquid.
    In this problem, volume of the displaced liquid is the volume of the cylindrical portion of the glass tube which is inside the water.
    This volume V = volume of the solid cylinder of radius ( r + d ) - volume of the solid cylinder of radius r. That is given in the above equation.
     
  13. Aug 1, 2009 #12
    OK.What if we were asked to find the upthrust acting on an object,say a cubed shaped metal object with a cavity,then wouldn't we take the volume of the metal part of the object+volume of the air cavity,immersed in the fluid?
     
  14. Aug 1, 2009 #13

    rl.bhat

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    Only outer volume of the cube.
     
  15. Aug 1, 2009 #14
    Isn't that the same as taking the volume of the metal part+volume of the cavity?I'm sorry my English is not very good,I tend to misunderstand things easily,so I apologise if I'm being annoying.

    If a question states the volume of the metal=x and the volume of the hole in the object as 'y',then to find the upthrust would we not take the total volume of the metal+hole which is 'x+y' ?
     
  16. Aug 1, 2009 #15

    rl.bhat

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    The solid cube of x volume and a cube x volume with a hole of volume y, will have the same upthrust. In fact a sold sphere of radius R and a hallow sphere of the same radius will have the same upthrust.
     
  17. Aug 2, 2009 #16
    OK,then in this problem wouldn't the upthrust acting on this tube be the same as the upthrust acting on a solid rod,i.e U=pi*(r+d)^2*h*density*g ?or is it different cause it's not completely immersed in the liquid,which is why U=π[( r + d)^2 - r^2]*h?

    Already grateful for all the help.
     
  18. Aug 2, 2009 #17
    Leena, rl.bhats method is far far better than mine.My method applies only to the special case where r equals zero and it is a solid rod whereas rl.bhats proof is general and shows that the answer is independant of r.

    Well done rl.bhat.Like yourself my initial assumption was that r was needed but unlike yourself I didn't pick up my pen and work it out properly. :blushing:

    Leena ...a few notes of clarification:
    1.The volume of water displaced includes that of the cavity but only if it were sealed and water couldn't enter.Imagine a ship getting holed and taking on water.
    2.With a tube, surface tension pulls down on the inside as well as the outside and hence the equation that rl.bhat wrote.
     
    Last edited: Aug 2, 2009
  19. Aug 3, 2009 #18
    Sorry.I'm not sure I understand this. If the cavity was sealed in then why should we add the volume of the cavity to find the upthrust?
    Shouldn't we just take the outer volume of the object without the cavity?
    Isn't this what rl.bhat says in posts #11 & #13 ?(at least that's what I think,please correct me if I'm wrong)

    The reason I'm so confused is,we did another question in class based on the Archimedes' principle ,where we were asked to find the volume of the cavity(the cavity here allows water to completely flow through)in an object.

    Here's the question:
    A metal object of mass 8kg with a cavity is attached to an inflated spherical rubber balloon with a light string,as shown in the figure.
    [URL=http://img15.imageshack.us/my.php?image=11398294.png][PLAIN]http://img15.imageshack.us/img15/2162/11398294.th.png[/URL][/PLAIN]

    When the radius of the balloon is 10cm,the system just floats in the deep lake .
    Density of metal(d) is 8000kgm-3 and the density of water (rho)is 1000kgm-3

    1)Neglecting the mass of the balloon find the volume of the cavity in the metal object.

    The solution :
    Volume of metal object with cavity=V'
    Volume of balloon=V"

    Law of flotation

    Mg = (V'+V")*rho*g
    8g = (V+ 4/3*pi*r3)*rho
    Then we found V' which is equal to 3.81*10-3 m3

    Volume of metal only = mass/density
    =8/8000=1*10-3m3

    Then to find the volume of cavity
    V' = Volume of metal + volume of cavity
    3.81*10-3=1*10-3 + V
    2.81*10-3m3 = V

    My question is
    why in this question,we have taken the volume of the cavity as well as the volume of the metal to find the upthrust,even though water flows through the hole?
    And in the surface tension question on the tube ,we've taken only the volume of the solid part of it and not the volume of the hole or cavity?
    To me, both methods seem to contradict eachother,yet they both yield the correct answer.
    But I don't understand how it works and why we include the volume of the cavity in one and in the other we completely ignore it?


    I have looked everywhere for an explanation,in the PF library(found something on buoyant force,which explained the buoyant force in solid objects well but was not much related to this question I think)and countless websites,.still couldn't find anything on the upthrust acting on objects with cavities.

    I've always had problems with fluid dynamics,so please help me understand this.
    I hope someone could at least provide me some links on the Archimedes' principle?

    Any and all help is much appreciated.
     
  20. Aug 3, 2009 #19
    Hello leena ,
    In the case of the capillary tube we do just consider the outer volume of the object but if it was sealed, so that water could not enter, then we would have to take the whole submerged volume into account.Imagine pushing a pipe ,which was sealed at the bottom with a cork,under water.This would cause the water level outside to rise.If we now do the same thing without the cork then less volume would be displaced and the rise would not be as high.
    With the Archimedes problem and the solution given the metal must be hollow and completely sealed so that no water can enter.There are no holes, even thought the diagram and the word "cavity" imply that there are.If holes were punched in the metal and water entered then the upthrust would get smaller and the whole thing will sink.
     
  21. Aug 3, 2009 #20
    Yes.That makes sense.
    It's just that we had done a couple f more similar questions on the Archimedes' principle and we had always used the same method i.e. taken the whole volume of the object.And all of the diagrams given looked like the cavity would go right through the body and so I just assumed it was so.My mistake.

    But I realise my mistake now.All thanks to you and rl.bhat!
    Thank you so much!
     
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