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anonymous
 4 years ago
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?
anonymous
 4 years ago
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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@tanvidais13 @csaltos @uncle

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0whenever 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hey i hav to find percentage error intape.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it will be change in length/original length*100

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0bt length is nt mention only temp. is mention

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0L in this formula will be cancelled as it is in both numerator and denominator..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hey cn u solve it i'm nt getting u?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait for 5 minute then i will upload the solution

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay tell me where u have the doubt?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0orignal lenth is calibrated length and change is occur when measured at 15c

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hey ash can u hlp me out

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.2Yes. Where do you have trouble understanding?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i'm nt geting the percentage error

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.2All 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
 4 years ago
Best ResponseYou've already chosen the best response.0yes,after put result answer is nt coming

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.2Let's find the change in length \(\alpha \) for steel= \(10.8\times 10^{6} / C\) C= degree centigrade What's \(\Delta T\) here?

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.2it's 35. The length will decrease \[\Delta T=(15)(20)=35\]

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.2\[\text{now find}\ \Delta L \ \text{in terms of L} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0nt getting ANSWER CORRECT

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.2Do you have the answer?

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.2Does you question mentions the value of alpha?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no ur answer that u get

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.2It's because of the value of alpha. For steel it varies between \(1113 *10^{(6)} /C\) Use \(12\times 10^{6} /C\) and you'd get the answer

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i think that if we use difference in reading we get the answer.in place of length

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@sauravshakya @shadowfiend @TuringTest
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