Two lengths x and y which are measured in an experiment are
\[x= ( 10.0 \pm 0.1 ) cm\] and\[y= (5.0\pm0.1)cm\] Calculate the uncertainty in a) (x+y) b)(x-y) c)xy d) \[\frac{ x }{ y }\]
@Somy
alright look you need to memorize this In addition or subtraction= errors are added up In multiplication or division= percentage errors are added up
okay..
x= 10.0 +- 0.1 y= 5.0 +- 0.1
a) 10+5= 15 0.1+0.1 =0.2 so 15+-0.2
uncertainty in A will be 0.2
same for B
b) 10-5=5 0.1+0.1=0.2
yes thats right :) the q is asking for uncertainty alone so it'll be 0.2
now C
do u know how to find %error?
% error= (error/actual value) *100
get it for both x and y
can u show me show me the working for c)
ok % error in X= (0.1/10)* 100= 1% % error in Y= (0.1/5)*100= 2% so in multiplication or division % errors are added up so % error= 1+2= 3% now C) xy = 10*5= 50 50----100% X-----3% X= 50*3/100 X= 1.5 thus uncertainty in C is 1.5
get it?
Trying to undertand...
alright look the value given for X= 10.0 +-0.1 right?
yes
0.1 is error right?
ys
ok so to find % error we have a formula %error= error/actual value *100= actual value in case of X is 10 get it?
ys
now whatever type of calculation that it is addition, subtraction, multiplication or division ACTUAL values do them
meaning 10 in case of X and 5 in case of Y
clear?
ys
the errors of X and Y do not take part in actual calculation meaning they don't take part in addition, subtraction, multiplication or division
errors have their own rule When its addition or subtraction between the actual values errors of actual values will ALWAYS ADD UP
ok
now when its multiplication or division between actual values PERCENTAGE ERRORS will ALWAYS ADD UP
get it?
yes
alright so do u understand the calculation now?
ys for d) its 0.06 is it so
when you find % error, and now want to find uncertainty what you do is that you take a product of multiplication or division of actual values as 100% and uncertainty is % error that you found
Okay
yup that's right)
Thanks for the explanation @Somy
you are welcome :)
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