## anonymous 3 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?

1. anonymous

@tanvidais13 @csaltos @uncle

2. anonymous

@HELLSGUARDIAN

3. 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?

4. anonymous

hey i hav to find percentage error intape.

5. anonymous

it will be change in length/original length*100

6. anonymous

bt length is nt mention only temp. is mention

7. anonymous

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

8. anonymous

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

9. anonymous

wait for 5 minute then i will upload the solution

10. anonymous

okay

11. anonymous

12. anonymous

you get it?

13. anonymous

no

14. anonymous

okay tell me where u have the doubt?

15. anonymous

change in length

16. anonymous

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

17. anonymous

have u got it

18. anonymous

@UnkleRhaukus

19. anonymous

hey ash can u hlp me out

20. ash2326

Yes. Where do you have trouble understanding?

21. anonymous

i'm nt geting the percentage error

22. 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?

23. anonymous

yes,after put result answer is nt coming

24. 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?

25. anonymous

35

26. ash2326

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

27. ash2326

$\text{now find}\ \Delta L \ \text{in terms of L}$

28. anonymous

29. ash2326

30. anonymous

YES

31. ash2326

Mind sharing here?

32. anonymous

its -0.042%

33. ash2326

Does you question mentions the value of alpha?

34. anonymous

no ur answer that u get

35. 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

36. ash2326

Do you get it?

37. anonymous

99.94%

38. anonymous

no .

39. anonymous

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

40. anonymous