Question

A screw has pitch equal to 0.5mm. What should be the number of divisions on its head so as to read correctly up to $0.001 mm$ with its help?

Answer 1

Hint: Pitch is the distance moved by the head of the screw gauge in one complete revolution. Use the formula for least count in terms of pitch and number of divisions on the main scale.

Formula used:
Least count of a measuring instrument is given as,
\[{\text{L}}{\text{.C}} = \dfrac{{{\text{Pitch}}}}{{{\text{Number of divisions}}}}\]

Complete step by step solution:
We have given that the pitch of the measuring screw is 0.5 mm and also the least count of that screw should be 0.001 mm. We know the least count of the measuring instrument is the ratio of the pitch of the measuring instrument to the number of divisions on its main scale. That is,
\[{\text{L}}{\text{.C}} = \dfrac{{{\text{Pitch}}}}{{{\text{Number of divisions}}}}\]
\[ \Rightarrow {\text{Number of divisions}} = \dfrac{{{\text{Pitch}}}}{{{\text{L}}{\text{.C}}}}\]
Substituting the given quantities in the above equation, we get,
\[{\text{Number of divisions}} = \dfrac{{{\text{0}}{\text{.5}}\,{\text{mm}}}}{{{\text{0}}{\text{.001}}\,{\text{mm}}}}\]
\[ \therefore {\text{Number of divisions}} = {\text{500}}\]

Therefore, the screw gauge should have 500 number divisions on its head to measure correctly up to 0.001 mm.

Additional information:
Pitch is the distance moved by the head of the screw gauge in one complete revolution. If we want to determine the pitch of the measuring instrument, it is the ratio of the distance moved by the spindle to the number of given rotations.The least count of some of the measuring instruments is,The least count of steel scale is 1 mm, least count of vernier caliper is 0.02 mm and least count of micrometer screw gauge is 0.001 mm.

Note:Least count is a measure of accuracy but it does not solely mean accuracy. To use the formula for least count make sure that value of pitch is given in $mm$ if you want least count in mm. Always express the least count of the measurement instrument in mm instead of $cm$ since it represents the smallest possible measurement.