Question

A screw gauge advances by$3mm$ on the main scale in $6$rotations. There are$50$ divisions on a circular scale. Find the least count of screw gauges.

• A. $0.01cm$
• B. $0.001cm$
• C. $0.001mm$
• D. $0.02mm$

Answer 1

The correct option is B

$0.001cm$

Step 1. Given data :

Main scale = $3mm$

Rotations = $6$

Divisions = $50$

Step 2. Find the pitch :

Pitch = (main scale reading) / (no. of rotations given).

$$\begin{gathered} =\frac{3}{6} \\ =0.5mm \end{gathered}$$

Step 3. Find the least count of screw gauge:

Least count = Pitch / number of divisions

$=\frac{0.5}{50}$

$=0.01mm$

$=0.001cm$

Hence, option (B) is correct.