A Vernier calliper has a least count of 0.02 mm and each main scale...

The least count (LC) of a Vernier calliper is given by the difference between one main scale division and one Vernier scale division.

$\text{LC} = 1\text{ MSD} - 1\text{ VSD}$

Given that $\text{LC} = 0.02\text{ mm}$ and $1\text{ MSD} = 0.5\text{ mm}$, we can write:
$0.02\text{ mm} = 0.5\text{ mm} - 1\text{ VSD}$

Solving for $1\text{ VSD}$:
$1\text{ VSD} = 0.5\text{ mm} - 0.02\text{ mm} = 0.48\text{ mm}$

To find the coincidence ratio, we take the ratio of the length of a VSD to an MSD:
$\dfrac{1\text{ VSD}}{1\text{ MSD}} = \dfrac{0.48}{0.50} = \dfrac{48}{50} = \dfrac{24}{25}$

Rearranging this gives:
$25\text{ VSD} = 24\text{ MSD}$

Thus, $25$ Vernier scale divisions coincide with $24$ main scale divisions.

Answer: $25\text{ VSD}$ coincide with $24\text{ MSD}$