First slide

Accuracy; precision of instruments and errors in measurements

Question

The diameter of a wire as measured by screw gauge was found to be 2.625, 2.630, 2.628 and 2.626 cm. Calculate mean absolute error in each measurement.

Moderate

A

0.02 cm
B

0.002 cm
C

0.001 cm
D

0.0002 cm

Solution

$a_{m} = \frac{2.625 + 2.630 + 2.628 + 2.626}{4} = 2.627 cm$
Taking $a_{m}$ as the true value, the absolute errors in different observations are,
$Δ a_{1} = 2.627 - 2.625 = + 0.002 cm$
$Δ a_{2} = 2.627 - 2.630 = - 0.003 cm$
$Δ a_{3} = 2.627 - 2.628 = - 0.001 cm$
$Δ a_{4} = 2.627 - 2.626 = 0.001 cm$
$\bar{△ a} = \frac{0.007}{4} = 0.00175 cm = 0.002 c m$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App