Question

The thimble of a screw gauge has $100$divisions. The spindle advances$2mm$ when the screw is turned through four revolutions. What is the least count of the screw gauge?

• A. $0.0005cm$
• B. $0.005cm$
• C. $0.050m$
• D. $0.050mm$

Answer 1

The correct option is A

$0.0005cm$

Step 1: Given data and formula used

$Pitch=\frac{(movingdis\tan cebyascrew\left(x\right))}{(numberofrotations\left(n\right))}=\frac{2mm}{4}=0.5mm$

Pitch(P) = 0.5 mm

Number of head scale division(N) = 50

$(L.C)$stands for least count

$L\mathit{.}C\mathit{=}\frac{P}{N}$

Step 2: Calculation of least count

$L\mathit{.}C\mathit{=}\frac{P}{N}\mathit{=}\frac{\mathit{0}\mathit{.}\mathit{5}}{\mathit{100}}\mathit{=}\mathit{0}\mathit{.}\mathit{005}\mathit{}mm\mathit{=}\mathit{0}\mathit{.}\mathit{0005}\mathit{}cm$

Hence, the correct option is (a).