The diameter of a wire measured in an experiment was 0.022cm, 0.023cm, 0.026cm, 0.025cm, 0.024cm" and "0.025cm find (i) the mean value of diameter (ii) absolute error in a each measurment (iii) percentage error.

The diameter of a wire measured in an experiment was 0.022cm, 0.023cm,

1.	The diameter of a wire measured in an experiment was 0.022cm, 0.023cm, 0.026cm, 0.025cm, 0.024cm" and "0.025cm find (i) the mean value of diameter (ii) absolute error in a each measurment (iii) percentage error.
Answer» Solution :(i) Mean DIAMETER `d_(m)=(0.022+0.023+0.026+0.025+0.024+0.025)/(6)` `=0.02416 cm` `d_(m)= 0.024cm` [Rounded off to 2 significant figures] (II) Absolute errors in the different measurment are `triangled_(1)= d_(m)-d_(1)` `=0.024-0.022 = 0.002cm` `triangled_(2)= d_(m)-d_(2)` `=0.024-0.023= 0.001cm` `triangled_(3)= d_(m)-d_(3)` `=0.024-0.026= -0.002cm` `triangled_(4)= d_(m)-d_(4)` `=0.024-0.025= -0.001cm` `triangled_(5)= d_(m)-d_(5)` `=0.024-0.024=0` `triangled_(6)=d_(m)-d_(6)` `=0.024-0.025= -0.001cm`. (iii) Absolute errors `=(sum\|triangled_(i)\|)/(6)` `=(0.002+0.001+0.002+0.001+0.001+0)/(6)` `=0.0012=0.001` (Rounded off to 1 significiant figure) PERCENTAGE error `=("Absolute error")/(d_m)xx100` `=(0.001)/(0.024)xx100` `=4.2%= 4%` (Rounded off to 1 significant figure).