Question

The least count of a micrometer is \[0.001\,cm\]. Using it the diameter of a wire is measured as \[0.084\,cm\]. Relative error in this measurement is:
A. \[\dfrac{{10}}{{84}}\]
B. \[\dfrac{1}{{840}}\]
C. \[\dfrac{1}{{84}}\]
D. None

Answer 1

Hint: Use the formula for relative error in screw gauge. Relative error in a screw gauge or micrometer is given by the ratio of least count to the measured value.Screw Gauge is a mechanical tool which facilitates measuring diameter or radius or thickness of thin wire or of metal sheets with accuracy.

Formula used:
\[R.E = \dfrac{{l.c}}{d}\]
where, \[R.E\] is a relative error, \[l.c\] is the least count of the micrometer and \[d\] is the measured value.

Complete step by step answer:
We know that the relative error in a screw gauge or micrometer is given by the ratio of least count to the measured value.We have, \[l.c = 0.001\,cm\] and the measured value of the diameter is given as,
\[R.E = \dfrac{{l.c}}{d}\]
Hence, relative error in the measurement will be,
\[R.E = \dfrac{{0.001}}{{0.084}} \]
\[\Rightarrow R.E= \dfrac{1}{{84}}\]
So, relative error is \[\dfrac{1}{{84}}\]

Hence, option C is correct.

 Note: Relative error in a measurement is the ratio of the absolute error of a measurement to the measurement being taken. Here, absolute error for the equipment is the least count of the micrometer screw. Since, this is the minimum value that the micrometer can measure. So, absolute error will be the same as the least count of the device. Percentage of error of the device is calculated as, \[P.E = \dfrac{{l.c}}{d} \times 100\% \] where, \[P.E\] is the percentage error.