### Related Formula The Least Count (LC) of a screw gauge tool is defined by: textL.C. = fractextPitchtextTotal Number of Circular Divisions (N) $\text{L.C.} = \frac{\text{Pitch}}{\text{Total Number of Circular Divisions (N)}} $ ### Core Logic The initial least count is given as [cite: 167, 788]: textL.C.textinitial = fracPN = 0.01text mm$\text{L.C.}{\text{initial}} = \frac{P}{N} = 0.01\text{ mm}$ Now, calculate the modified parameters from the text details : * New Pitch: P' = P(1 + 0.75) = 1.75P$P' = P(1 + 0.75) = 1.75P$ * New Divisions: N' = N(1 - 0.50) = 0.5N$N' = N(1 - 0.50) = 0.5N$ ### Step 1: Calculating the New Least Count Set up the updated least count expression ratio : textL.C.textnew = fracP'N' = frac1.75P0.5N = 3.5 times left(fracPN ight)$\text{L.C.}{\text{new}} = \frac{P'}{N'} = \frac{1.75P}{0.5N} = 3.5 \times \left(\frac{P}{N} ight)$ Substitute the initial least count value : textL.C.textnew = 3.5 times 0.01text mm = 0.035text mm$\text{L.C.}{\text{new}} = 3.5 \times 0.01\text{ mm} = 0.035\text{ mm}$ Converting into the requested scientific prefix units (10^-3text mm$10^{-3}\text{ mm}$) : textL.C.textnew = 35 times 10^-3text mm$\text{L.C.}{\text{new}} = 35 \times 10^{-3}\text{ mm}$ Therefore, the requested value is 35. ### Pattern Recognition Least count scales proportionally with pitch increases, and inversely with reductions in circular divisions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements