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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

okay

you get it?

no

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

yes,after put result answer is nt coming

35

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

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

nt getting ANSWER CORRECT

Do you have the answer?

YES

Mind sharing here?

its -0.042%

Does you question mentions the value of alpha?

no
ur answer that u get

Do you get it?

99.94%

no .

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