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Q.

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.

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a

0.02 cm

b

0.002 cm

c

0.001 cm

d

0.0002 cm

answer is B.

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Detailed Solution

am=2.625+2.630+2.628+2.6264=2.627 cmTaking am as the true value, the absolute errors in different observations are,Δa1=2.627-2.625=+0.002 cmΔa2=2.627-2.630=-0.003 cmΔa3=2.627-2.628=-0.001 cmΔa4=2.627-2.626=0.001 cm△a¯=0.0074=0.00175 cm = 0.002 cm
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