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

Input Parameters:

- Original least count: \( 0.01 \, \text{mm} \)

- Pitch increase: 75% (New pitch = \( 1.75 \times \text{Original pitch} \))

- Circular scale divisions reduction: 50% (New divisions = \( \frac{1}{2} \times \text{Original divisions} \))

Step 1: Least Count Formula

The formula for least count \( LC \) of a screw gauge is: \[ LC = \frac{\text{Pitch}}{\text{Number of divisions on the circular scale}}. \]

Step 2: Incorporate Modifications

With the new pitch \( P_{\text{new}} = 1.75 \times P \) and new divisions \( N_{\text{new}} = \frac{N}{2} \), the new least count is calculated as: \[ LC_{\text{new}} = \frac{1.75 \times P \times 2}{N} = 3.5 \times LC_{\text{original}}. \]

Step 3: Calculate New Least Count

Using the original least count \( LC_{\text{original}} = 0.01 \, \text{mm} \), the new least count is: \[ LC_{\text{new}} = 3.5 \times 0.01 = 0.035 \, \text{mm}. \]

Final Result:

The updated least count is \( \boxed{0.035 \, \text{mm}} \).