The least count of the main scale of a screw gauge is 1 mm. The minimu

1.	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 is1. 502. 2003. 5004. 100
Answer» 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

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 is1. 502. 2003. 5004. 100

Answer» 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

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