Question 27: 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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