Question

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:

Hint:

The least count is inversely proportional to the number of divisions on the circular scale. An increase in pitch and a decrease in the number of divisions both affect the least count, so calculate the changes in pitch and divisions carefully.

Answer 1

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 \( \boxed{0.035 \, \text{mm}} \).