Question:
A screw gauge has pitch \(=0.5\ \text{mm}\) and number of divisions on circular scale \(=50\). Find least count.
A screw gauge has pitch \(=0.5\ \text{mm}\) and number of divisions on circular scale \(=50\). Find least count.
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For screw gauge:
\[
\text{Least Count}=\frac{\text{Pitch}}{\text{Number of circular scale divisions}}
\]
Updated On: Jun 3, 2026
- \(0.1\ \text{mm}\)
- \(0.01\ \text{mm}\)
- \(0.001\ \text{mm}\)
- \(0.05\ \text{mm}\)
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The Correct Option is B
Solution and Explanation
Concept:
Least count of a screw gauge is given by: \(\displaystyle \text{Least Count}=\frac{\text{Pitch}}{\text{Number of divisions on circular scale}}\)
Step 1: Write the given values.
Pitch: \(\displaystyle 0.5\ \text{mm}\) Number of circular scale divisions: \(\displaystyle 50\)
Step 2: Apply the formula.
\(\displaystyle \text{Least Count}=\frac{0.5}{50}\ \text{mm}\) \(\displaystyle =0.01\ \text{mm}\)
Step 3: Final conclusion.
Hence, the least count is: \(\displaystyle \boxed{0.01\ \text{mm}}\)
Least count of a screw gauge is given by: \(\displaystyle \text{Least Count}=\frac{\text{Pitch}}{\text{Number of divisions on circular scale}}\)
Step 1: Write the given values.
Pitch: \(\displaystyle 0.5\ \text{mm}\) Number of circular scale divisions: \(\displaystyle 50\)
Step 2: Apply the formula.
\(\displaystyle \text{Least Count}=\frac{0.5}{50}\ \text{mm}\) \(\displaystyle =0.01\ \text{mm}\)
Step 3: Final conclusion.
Hence, the least count is: \(\displaystyle \boxed{0.01\ \text{mm}}\)
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