A Vernier caliper has a least count of 0.02 mm . The main scale is...

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

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

It is given that $N$ divisions of the Vernier scale coincide with $(N-1)$ divisions of the main scale.

$N \times \text{VSD} = (N-1) \times \text{MSD}$

$1\text{ VSD} = \dfrac{N-1}{N}\text{ MSD}$

Substituting this into the least count formula:

$\text{LC} = 1\text{ MSD} - \left(\dfrac{N-1}{N}\right)\text{ MSD}$

$\text{LC} = \dfrac{1}{N}\text{ MSD}$

We are given that $\text{LC} = 0.02\text{ mm}$ and $1\text{ MSD} = 0.5\text{ mm}$.

$0.02\text{ mm} = \dfrac{0.5\text{ mm}}{N}$

$N = \dfrac{0.5}{0.02} = 25$

Therefore, the number of divisions on the Vernier scale is $25$.

Answer: $25$