Question 28: Units , Dimensions and Error Analysis

Question

The relative errors in the measurement of two lengths 1.02 cm $$\pm$$ 0.01 cm and 9.89 cm $$\pm$$ 0.01 cm is

Accepted Answer

The error in measurement of length 1.02 cm is $$\pm$$0.01 cm and error in measurement of length 9.89 cm is $$\pm$$0.01 cm.
$$	herefore$$ The relative error is 1.02 cm
$$ = \left[ { \pm {{0.01} \over {1.02}} 	imes 100\% } ight] =  \pm 0.98\%  =  \pm 1\% $$
Similarly, the relative error in 9.89 cm
$$ = \left[ { \pm {{0.01} \over {9.89}} 	imes 100\% } ight] =  \pm 0.1\% $$
Therefore, relative errors in measurement of two masses are $$\pm$$1% and $$\pm$$0.1%.

Answer

$$\pm$$1% and $$\pm$$0.1%

Answer

$$\pm$$1% and $$\pm$$0.2%

Answer

$$\pm$$1% and $$\pm$$1%

Answer

$$\pm$$0.1% and $$\pm$$1%

Units , Dimensions and Error Analysis

The relative errors in the measurement of two lengths 1.02 cm $\pm 0.01 cm and 9.89 cm \pm$ 0.01 cm is

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