Consider a Vernier callipers in which each `1cm` on the main scale is divided into `8` equal divisions and a screw gauge with `100` divisions in its circular scale. In the Vernier callipers, `5` divisions of the Vernier scale coincide with `4` divisions on the scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then :

Explanation

To solve the problem, we need to determine the least count of both the Vernier calipers and the screw gauge based on the information provided.

### Step-by-Step Solution:

1. **Identify the Main Scale Division (MSD) of the Vernier Calipers:**
- Each `1 cm` on the main scale is divided into `8` equal divisions.
- Therefore, the value of one Main Scale Division (MSD) is:
\[
...