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Thermal Physics

Blazing goIITian

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1 Jan 2008 22:09:37 IST

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thermal expansion problem

None

Q. At 0⁰ C , three metal rods form an equilateral triangle. Two rods are of the same material, but the third rod is made of Invar ( its expansion is negligible ). When the triangle is heated upto 100⁰ C, the angle formed between the two metal rods of the same material is ( pi/ 3 - theta ).
Prove that the co-efficient of linear expansion of the two metal rods is 3 theta / 200 per degree centigrade.

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Comments (6)

ramyani chakrabarty

Blazing goIITian

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1 Jan 2008 22:17:16 IST

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pointer assured

abhishek sinha

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1 Jan 2008 22:34:48 IST

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let the base be 2b, and lenth of each equal side =L

Then from geometry , L sin@ = b = const ( @ is the half vertical angle )

diff both sides L cos @ d @ + sin @ dL = 0 .............( 1)

again from defn = 1/L dL / dT

so dL = L dT .................( 2 )

substituting (2 ) in ( 1 ) , we get

= - cot @ d@/ dT

Now for small change , d@ = - /2 & dT = 100 & @ = 30

so we get = sqrt( 3) /200

( proved )

ramyani chakrabarty

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1 Jan 2008 22:54:06 IST

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will u plz explaim the lat two steps.

Now for small change , d@ = - /2 & dT = 100 & @ = 30

so we get = sqrt( 3) /200

Karthik M

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1 Jan 2008 23:38:50 IST

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Feynmann... your method is short and sweet... But I too didn't follow that last argument of yours...I think it must be d@ = -theta... anyways... here is my (longer) method...

When they are heated, since they are made of same material, both should expand uniformly.

So, (consider invar rod as base), the new base angle will be:

2

+ (60-@) = 180 or @ = 60-@/2.

Now, L'cos(60-@/2) = L. (L' is the new expanded length)

expanding cos(60-@/2), assuming that cos@/2 = 0 (as @ is small, and sin@/2 = @/2)

We get the LHS as L'(1/2 -

3/4) = 1/2.

now, L' = L(1+100

).

Putting this in the above equation, we get (1+100

) = (1-

3@/2)^-1

Using binomial, we get (1+100

) = 1+

3@/2

Or

3@/200

I think I had done this problem last year somewhere... not too sure though...

anyway... feynmann's method sure does seem much shorter.

ramyani chakrabarty

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2 Jan 2008 00:47:26 IST

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got it now!!! thnks karthik!!

abhishek sinha

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7 Jan 2008 12:38:30 IST

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Sorry for the delay .

Now see that @ is the semi vertical angle .

So @ = 60 / 2 = 30 deg

again the total vertical angle decreases by theta . So obviously the semi vertical angle decreases by theta / 2 .

so d ( @ ) = @ = - theta / 2

and dT = T = 100 deg C

Hope that it is clear now .

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