In a specially designed vernier calipers, 10 divisions of the...

Given $10\text{ VSD} = 11\text{ MSD}$.

$1\text{ VSD} = \dfrac{11}{10}\text{ MSD} = 1.1\text{ MSD}$.

Since $1\text{ MSD} = 1\text{ mm}$, we get $1\text{ VSD} = 1.1\text{ mm}$.

This is a retrograde vernier where the vernier division is larger than the main scale division. The least count is the magnitude of the difference between one vernier scale division and one main scale division.

Least count $= |1\text{ MSD} - 1\text{ VSD}|$

Least count $= |1.0\text{ mm} - 1.1\text{ mm}| = 0.1\text{ mm}$.

Answer: $0.1\text{ mm}$