Question

The screw advances by $0.5mm$ on two complete rotations and number of division on the circular scale is $125$. The least screw gauge is

• A. $0.002mm$
• B. $0.02cm$
• C. $0.02mm$
• D. $0.002m$

Answer 1

The correct option is A $0.002mm$

The least screw gauge is given by,

$=\frac{\frac{0.5}{2}}{125}$

$=\frac{0.25}{125}$

$=2\times {10}^{-3}$

$=0.002mm$