Question

The distance between two consecutive threads of the screw of a screw gauge is 1 mm and the least count of the screw gauge is 0.01 mm. What is the number of head scale divisions?

• A. $50$
• B. $100$
• C. $150$
• D. $200$

Answer 1

The correct option is B

$100$

Step 1: Given the data and formula used

$\mathrm{Pitch}\mathrm{of}\mathrm{the}\mathrm{screw}\mathrm{gauge}=\frac{(\mathrm{moving}\mathrm{distance}\mathrm{by}\mathrm{a}\mathrm{screw})}{(\mathrm{number}\mathrm{of}\mathrm{rotations})}=\frac{3\mathrm{mm}}{3}=1mm$

$$\begin{gathered} L\mathrm{east}\mathrm{cound}(\mathrm{L}.\mathrm{C})=0.01 \\ \mathrm{Formula}\mathrm{of}\mathrm{least}\mathrm{count}\mathrm{L}.\mathrm{C}=\frac{\mathrm{pitch}}{\mathrm{number}\mathrm{of}\mathrm{head}\mathrm{scale}\mathrm{divisions}} \\ \mathrm{number}\mathrm{of}\mathrm{head}\mathrm{scale}\mathrm{divisions}=\frac{\mathrm{pitch}}{\mathrm{L}.\mathrm{C}} \end{gathered}$$

Step 2: Finding the value of the number of head scale divisions

On putting given value in the formula

$\mathrm{N}=\frac{\mathit{1}\mathit{}\mathrm{mm}}{\mathit{0}\mathit{.}\mathit{01}\mathrm{mm}}=100$

Hence, the correct option is (b).