The main scale of a circular vernier instrument has N divisions per...

The main scale has $N$ divisions per degree. Therefore, the value of one main scale division (MSD) is:
$1 \text{ MSD} = \dfrac{1}{N} \text{ degrees}$

Since $1 \text{ degree} = 60 \text{ minutes}$, we can express $1 \text{ MSD}$ in minutes:
$1 \text{ MSD} = \dfrac{60}{N} \text{ minutes}$

It is given that $V$ divisions of the vernier scale coincide with $(V-1)$ divisions of the main scale:
$V \text{ VSD} = (V-1) \text{ MSD}$
$1 \text{ VSD} = \dfrac{V-1}{V} \text{ MSD}$

The least count (LC) is the difference between one MSD and one VSD:
$\text{LC} = 1 \text{ MSD} - 1 \text{ VSD}$
$\text{LC} = 1 \text{ MSD} - \dfrac{V-1}{V} \text{ MSD} = \dfrac{1}{V} \text{ MSD}$

Substituting the value of $1 \text{ MSD}$ in minutes:
$\text{LC} = \dfrac{1}{V} \times \left(\dfrac{60}{N}\right) \text{ minutes} = \dfrac{60}{NV} \text{ minutes}$

Answer: $\dfrac{60}{NV}$