The least count of a vernier callipers is given as x mm. It is...

The least count (LC) of a vernier callipers is defined as:
$\text{LC} = 1\text{ MSD} - 1\text{ VSD}$

Given the coincidence relation:
$N\text{ VSD} = (N-2)\text{ MSD}$
$1\text{ VSD} = \dfrac{N-2}{N}\text{ MSD}$

Substitute this into the least count formula:
$\text{LC} = 1\text{ MSD} - \left(\dfrac{N-2}{N}\right)\text{ MSD}$
$\text{LC} = \left(1 - \dfrac{N-2}{N}\right)\text{ MSD}$
$\text{LC} = \left(\dfrac{N - N + 2}{N}\right)\text{ MSD} = \dfrac{2}{N}\text{ MSD}$

We are given that $\text{LC} = x$. Therefore:
$x = \dfrac{2}{N}\text{ MSD}$

Rearranging to find the value of $1\text{ MSD}$:
$1\text{ MSD} = \dfrac{Nx}{2}$

Answer: $\dfrac{Nx}{2}$