The least count of the main scale of a screw gauge is 1 mm. The minimum number of divisions on its circular scale required to measure 5 μm diameter of a wire is
1. 50
2. 200
3. 500
4. 100
1. 50
2. 200
3. 500
4. 100
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Correct Answer – Option 2 : 200
Concept:
In a screw gauge,
\({\rm{Least\;count}} = \frac{{{\rm{\;Measure\;of\;}}1{\rm{\;main\;scale\;division\;(MSD)\;}}}}{{{\rm{\;Number\;of\;division\;on\;circular\;scale\;}}}}\)
Calculation:
Here, minimum value to be measured/least count is 5 μm = 5 × 10-6 m
From the values given in the question,
\(\Rightarrow 5 \times {10^{ – 6}} = \frac{{1 \times {{10}^{ – 3}}}}{{\rm{N}}}\)
\(\Rightarrow {\rm{N}} = \frac{{{{10}^{ – 3}}}}{{5 \times {{10}^{ – 6}}}} = \frac{{1000}}{5}\)
∴ N = 200 divisions