Linear Expansion Homework: 40 & 40.1cm Markings Coincide

[leena19]

Homework Statement

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)

Homework Equations

l = l0 (1+ alpha*theta)

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.

 

[kuruman]

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?

 

[leena19]

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?

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.

 

[kuruman]

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?

 

[leena19]

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?

 

[kuruman]

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.

 

[leena19]

I do not advocate working back from a known answer. This is what I would say to compete 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.

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.