Question

The pitch and the number of circular scale divisions in a screw gauge with least count $0.02mm$ are respectively
A. $1mm$ and $100$
B. $0.5mm$ and $50$
C. $1mm$ and $50$
D. $0.5mm$ and $100$
E. $1mm$ and $200$

Answer 1

Least count is ratio of pitch and number of circular scale divisions. In this case the given quantity is the only least count. It can be solved only by a hit and trial method which needs to be applied for all options.

Formula Used: Least Count ${\text{ = }}\dfrac{{{\text{Pitch}}}}{{{\text{Number}}\,\,{\text{of}}\,\,{\text{circular}}\,\,{\text{scale}}\,\,{\text{division}}}}$

Complete step by step answer:
In this question, the only thing provided is the value of least count. As such we cannot calculate pitch and number of circular scale divisions. Only way is to check the ratio of all options.
Option A: $\dfrac{{1mm}}{{100}} = \,0.01mm$[does not match]
Option B: $\dfrac{{0.5mm}}{{50}} = \,0.01mm$[does not match]
Option C: $\dfrac{{1mm}}{{50}} = \,0.02mm$[matches with given value]
Option D: $\dfrac{{0.5mm}}{{100}} = \,0.005mm$[does not match]
Option E: $\dfrac{{1mm}}{{200}} = \,0.005mm$[does not match]

So, the correct answer is “Option C”.

Additional Information:
In the given question, if I increase the circular division from $50$to$100,$ then the least count will become $\dfrac{{1mm}}{{100}} = 0.01mm.$ This means the device which could earlier measure length as small as $0.02mm$ will now be able to measure $0.01mm.$

Note:
Lesser the value of least count, better is the accuracy of the device.
We can increase the accuracy of any device by increasing the number of divisions on a circular scale.