The least count of a screw gauge is 0.01 mm. If the pitch is increased by 75% and the number of divisions on the circular scale is reduced by 50%, the new least count will be:
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Solution and Explanation
Given:
- Original least count = \( 0.01 \, \text{mm} \) - The pitch is increased by 75% (new pitch = \( 1.75 \times \text{original pitch} \)) - The number of divisions on the circular scale is reduced by 50% (new divisions = \( \frac{1}{2} \) of the original divisions)
Step 1: Formula for Least Count
The least count \( LC \) of a screw gauge is given by: \[ LC = \frac{\text{Pitch}}{\text{Number of divisions on the circular scale}}. \]
Step 2: Apply the Changes
The new pitch is \( P_{\text{new}} = 1.75 \times P \), and the new number of divisions is \( N_{\text{new}} = \frac{N}{2} \). Hence, the new least count is: \[ LC_{\text{new}} = \frac{1.75 \times P \times 2}{N} = 3.5 \times LC_{\text{original}}. \]
Step 3: Final Calculation
Substituting \( LC_{\text{original}} = 0.01 \, \text{mm} \), we get: \[ LC_{\text{new}} = 3.5 \times 0.01 = 0.035 \, \text{mm}. \]
Final Answer:
The new least count is \( \boxed{0.035 \, \text{mm}} \).
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