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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?

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whenever something is heated it expands & when cooled it 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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Other answers:

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

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