First slide

Accuracy; precision of instruments and errors in measurements

Question

The average speed of a train is measured by 5 students. The results of measurements are given below

Moderate

A

2.6%
B

3.5%
C

4.5%
D

5.5%

Solution

Mean value, $v_{m} = \frac{10.2 + 10.4 + 9.8 + 10.6 + 10.8}{5}$
$= \frac{51.8}{5} = 10.4 m s^{- 1}$
$Δ v_{1} = v_{m} - v_{1} = 10.4 - 10.2 = 0.2$
$Δ v_{2} = v_{m} - v_{2} = 10.4 - 10.4 = 0.0$
$Δ v_{3} = v_{m} - v_{3} = 10.4 - 9.8 = 0.6$
$Δ v_{4} = v_{m} - v_{4} = 10.4 - 10.6 = - 0.2$
$Δ v_{5} = v_{m} - v_{5} = 10.4 - 10.8 = - 0.4$
Mean absolute error,
$\bar{Δ v} = \frac{| Δ v_{1} | + | Δ v_{2} | + | Δ v_{3} | + | Δ v_{4} | + | Δ v_{5} |}{5}$
$= \frac{0.2 + 0.0 + 0.6 + 0.2 + 0.4}{5} = \frac{1.4}{5} = 0.28 m s^{- 1}$
Relative error $= \pm \frac{\bar{Δ v}}{v_{m}} = \pm \frac{0.28}{10.4}$
Percentage error $= \pm \frac{\bar{Δ v}}{v_{m}} \times 100 = \pm \frac{0.28}{10.4} \times 100 = \pm 2.6 %$

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