Question

a mocrometer screw gauge was used to measure the diameter of ball bearing. three measurements were found as follows;5.26mm,5.21mm,5.24mm. calculate the absolute error in the sum of these values

Answer 1

Answer on Question #57931-Physics-Mechanics

A micrometer screw gauge was used to measure the diameter of ball bearing. Three measurements were found as follows: 5.26mm, 5.21mm, 5.24 mm. Calculate the absolute error in the sum of these values

Solution

Usually the mean value $\bar{d}$ is taken as the true value. So,

$$\bar{d} = \frac{5.26 + 5.21 + 5.24}{3} = 5.237 \text{ mm}$$ $$\Delta d_1 = |d_1 - \bar{d}| = |5.26 - 5.237| = 0.023 \text{ mm}$$ $$\Delta d_2 = |d_2 - \bar{d}| = |5.21 - 5.237| = 0.027 \text{ mm}$$ $$\Delta d_3 = |d_3 - \bar{d}| = |5.24 - 5.237| = 0.003 \text{ mm}$$

The absolute error in the sum of these values is

$$\Delta(d_1 + d_2 + d_3) = \Delta(d_1) + \Delta(d_2) + \Delta(d_3) = 0.023 + 0.027 + 0.003 = 0.053 \text{ mm}.$$

Answer: 0.053 mm.

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