anonymous
  • anonymous
a steel tape is calibrated at 20'c .on a cold day when the temperature is -15'c ,what will be percentage error in tape?
Physics
schrodinger
  • schrodinger
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anonymous
  • anonymous
anonymous
  • anonymous
anonymous
  • anonymous
whenever something is heated it expands & when cooled it contracts..so this case can be considered as the case of cooling..use equation \[change \in length=- \alpha \Delta T L\] where L is original length ,\[\Delta T\] is change in temperature..and \[\alpha \] is coefficient of thermal expansion of steel.. when u have found the change in length the %ge error will be given as (change in length/original length)*100 can u do the calculations?

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anonymous
  • anonymous
hey i hav to find percentage error intape.
anonymous
  • anonymous
it will be change in length/original length*100
anonymous
  • anonymous
bt length is nt mention only temp. is mention
anonymous
  • anonymous
L in this formula will be cancelled as it is in both numerator and denominator..
anonymous
  • anonymous
hey cn u solve it i'm nt getting u?
anonymous
  • anonymous
wait for 5 minute then i will upload the solution
anonymous
  • anonymous
okay
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
you get it?
anonymous
  • anonymous
no
anonymous
  • anonymous
okay tell me where u have the doubt?
anonymous
  • anonymous
change in length
anonymous
  • anonymous
orignal lenth is calibrated length and change is occur when measured at -15c
anonymous
  • anonymous
have u got it
anonymous
  • anonymous
anonymous
  • anonymous
hey ash can u hlp me out
ash2326
  • ash2326
Yes. Where do you have trouble understanding?
anonymous
  • anonymous
i'm nt geting the percentage error
ash2326
  • ash2326
All materials' length changes with temperature. Change in length is given as \[\Delta L=\alpha \Delta T L\] Here \(\Delta T\) is the change in temp. \(\alpha\) is the linear coefficient of thermal expansion. Do you get this formula?
anonymous
  • anonymous
yes,after put result answer is nt coming
ash2326
  • ash2326
Let's find the change in length \(\alpha \) for steel= \(10.8\times 10^{-6} / C\) C= degree centigrade What's \(\Delta T\) here?
anonymous
  • anonymous
35
ash2326
  • ash2326
it's -35. The length will decrease \[\Delta T=(-15)-(20)=-35\]
ash2326
  • ash2326
\[\text{now find}\ \Delta L \ \text{in terms of L} \]
anonymous
  • anonymous
nt getting ANSWER CORRECT
ash2326
  • ash2326
Do you have the answer?
anonymous
  • anonymous
YES
ash2326
  • ash2326
Mind sharing here?
anonymous
  • anonymous
its -0.042%
ash2326
  • ash2326
Does you question mentions the value of alpha?
anonymous
  • anonymous
no ur answer that u get
ash2326
  • ash2326
It's because of the value of alpha. For steel it varies between \(11-13 *10^{(-6)} /C\) Use \(12\times 10^{-6} /C\) and you'd get the answer
ash2326
  • ash2326
Do you get it?
anonymous
  • anonymous
99.94%
anonymous
  • anonymous
no .
anonymous
  • anonymous
i think that if we use difference in reading we get the answer.in place of length
anonymous
  • anonymous
anonymous
  • anonymous

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