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

The least count of a screw gauge is given by the formula: \[ \text{Least Count} = \frac{\text{Pitch}}{\text{Number of divisions on the circular scale}}. \] Let the original pitch be \( P \) and the number of divisions on the circular scale be \( N \). Thus, the original least count is: \[ \text{Original Least Count} = \frac{P}{N}. \] Given that the original least count is \( 0.01 \, \text{mm} \), we have: \[ \frac{P}{N} = 0.01. \] Step 1: The pitch is increased by 75%, so the new pitch \( P' \) becomes: \[ P' = P + 0.75P = 1.75P. \] Step 2: The number of divisions on the circular scale is reduced by 50%, so the new number of divisions \( N' \) becomes: \[ N' = \frac{N}{2}. \] Step 3: The new least count will be: \[ \text{New Least Count} = \frac{P'}{N'} = \frac{1.75P}{\frac{N}{2}} = \frac{1.75 \times 2P}{N} = \frac{3.5P}{N}. \] Step 4: Since \( \frac{P}{N} = 0.01 \, \text{mm} \), the new least count becomes: \[ \text{New Least Count} = 3.5 \times 0.01 = 0.035 \, \text{mm} = 35 \times 10^{-3} \, \text{mm}. \] Thus, the new least count is \( \boxed{35 \times 10^{-3} \, \text{mm}} \).