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Linear expansion

  1. Aug 14, 2009 #1
    1. The problem statement, all variables and given/known data
    Two metre rulers made of metal of linear expansivity 25*10-6/degrees C are calibrated at 0 deg celsius.
    One end of each of the metre rulers is fixed to a vertical wall and held side by side horizontally.
    One of the meter rulers is maintained at 0 deg.C and the other at 100 deg C,which two scale markings coincide with each other ?

    (ans: 40 & 40.1cm)


    2. Relevant equations

    l = l0 (1+ alpha*theta)

    3. The attempt at a solution

    For the metre ruler kept at 100deg.C ,
    l = l0 (1+ alpha*theta)
    l = 1 (1 + 25*10-6 *100)
    l = 1.0025 mm

    So when the meter ruler at 100 deg C reads 1mm,the actual reading is 1.0025mm.
    I'm not sure if what I've found is relevant to this question,even if it is I don't know how to proceed from here.

    Hope someone can help.
    Thanx.
     
  2. jcsd
  3. Aug 14, 2009 #2

    kuruman

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    The problem is asking you to find the first mm marking on the "hot" ruler that matches one on the "cold" ruler.

    As you say, 1 mm on cold is 1.0000 + 0.0025 mm on hot. Then
    2 mm on cold is 2.0000 + 2*0.0025 on hot
    3 mm on cold is 3.0000 + 3*0.0025 on hot

    Do you see what's going on?
     
  4. Aug 14, 2009 #3
    Umm... I think I do.
    But is there maybe a shorter method of finding it without having to calculate it that way for 100cm?
    I tried deriving a equation,but still no luck.

    Thx.
     
  5. Aug 14, 2009 #4

    kuruman

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    OK. Let's start over again. I was hoping you would see the pattern. Here we go

    1 mm on cold is 1.0000 + 1*0.0025 on hot. Is there a match? No.
    2 mm on cold is 2.0000 + 2*0.0025 on hot. Is there a match? No.
    3 mm on cold is 3.0000 + 3*0.0025 on hot. Is there a match? No.
    .........
    x mm on cold is x + x*0.0025 on hot.

    If x represents the mm mark on cold that first matches another mm mark on hot, what must x*0.0025 be in mm?
     
  6. Aug 14, 2009 #5
    OK.I think I found one.
    Working back from the answer..
    l = l0 (1+ alpha*theta)
    1*n = (n-1) (1 + alpha*theta)
    n = (n-1) (1.0025)
    therefore n=401mm and n-1=400mm

    Does this make sense?
    or is there another way of doing it?
     
  7. Aug 14, 2009 #6

    kuruman

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    I do not advocate working back from a known answer. This is what I would say to complete my previous post

    Let x = the first mm mark on cold that first matches another mm mark on hot (x is a dimensionless number)

    Then x*0.0025 mm = 1 mm

    x = 1 mm/(0.0025 mm) = 400

    Therefore the 400 mm mark on cold matches the 401 mm mark on hot. The equation basically counts how many 0.0025 increments are necessary to add up to 1 mm. :wink:
     
    Last edited: Aug 14, 2009
  8. Aug 14, 2009 #7
    Oh I see now.
    Thank you very much.
    and i'm sorry I didn't see your previous post when i made my reply in post#5.
     
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