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OpenStudy (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?

OpenStudy (anonymous):

@tanvidais13 @csaltos @uncle

OpenStudy (anonymous):

@HELLSGUARDIAN

OpenStudy (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?

OpenStudy (anonymous):

hey i hav to find percentage error intape.

OpenStudy (anonymous):

it will be change in length/original length*100

OpenStudy (anonymous):

bt length is nt mention only temp. is mention

OpenStudy (anonymous):

L in this formula will be cancelled as it is in both numerator and denominator..

OpenStudy (anonymous):

hey cn u solve it i'm nt getting u?

OpenStudy (anonymous):

wait for 5 minute then i will upload the solution

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

OpenStudy (anonymous):

you get it?

OpenStudy (anonymous):

no

OpenStudy (anonymous):

okay tell me where u have the doubt?

OpenStudy (anonymous):

change in length

OpenStudy (anonymous):

orignal lenth is calibrated length and change is occur when measured at -15c

OpenStudy (anonymous):

have u got it

OpenStudy (anonymous):

@UnkleRhaukus

OpenStudy (anonymous):

hey ash can u hlp me out

OpenStudy (ash2326):

Yes. Where do you have trouble understanding?

OpenStudy (anonymous):

i'm nt geting the percentage error

OpenStudy (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?

OpenStudy (anonymous):

yes,after put result answer is nt coming

OpenStudy (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?

OpenStudy (anonymous):

35

OpenStudy (ash2326):

it's -35. The length will decrease \[\Delta T=(-15)-(20)=-35\]

OpenStudy (ash2326):

\[\text{now find}\ \Delta L \ \text{in terms of L} \]

OpenStudy (anonymous):

nt getting ANSWER CORRECT

OpenStudy (ash2326):

Do you have the answer?

OpenStudy (anonymous):

YES

OpenStudy (ash2326):

Mind sharing here?

OpenStudy (anonymous):

its -0.042%

OpenStudy (ash2326):

Does you question mentions the value of alpha?

OpenStudy (anonymous):

no ur answer that u get

OpenStudy (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

OpenStudy (ash2326):

Do you get it?

OpenStudy (anonymous):

99.94%

OpenStudy (anonymous):

no .

OpenStudy (anonymous):

i think that if we use difference in reading we get the answer.in place of length

OpenStudy (anonymous):

@sauravshakya @shadowfiend @TuringTest

OpenStudy (anonymous):

@experimentX

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