Question

The diameter of a wire as measured by screw gauge was found to be $2.620,2.625,2.628and2.626cm$. Calculate absolute error in each measurement.

• A. $\Delta a_5=2.626-2.626=+0.000cm$
• B. $\Delta a_4=2.626-2.626=0.000cm$
• C. $\Delta a_3=2.626-2.630=-0.004cm$
• D. All of these

Answer 1

The correct option is B $\Delta a_4=2.626-2.626=0.000cm$

Given,

$a_1=2.620cm$

$a_2=2.625cm$

$a_3=2.628cm$

$a_4=2.626cm$

The mean value , $a_m=\frac{a_1+a_2+a_3+a_4}{4}$

$a_M=\frac{2.620+2.625+2.628+2.626}{4}$

$a_m=2.626cm$

Absolute error, $\Delta a_i=a_m-a_i$

$\Delta a_1=a_m-a_1=2.626-2.620=0.006cm$

$\Delta a_2=a_m-a_2=2.626-2.625=0.001cm$

$\Delta a_3=a_m-a_3=2.626-2.628=-0.002cm$

$\Delta a_4=a_m-a_4=2.626-2.626=0cm$