A specially designed Vernier caliper has a least count of 0.005 cm...

Let the length of one main scale division be $1\text{ MSD}$ and one Vernier scale division be $1\text{ VSD}$.

Given that $20\text{ VSD} = 19\text{ MSD}$, we have:
$1\text{ VSD} = \dfrac{19}{20}\text{ MSD}$

The least count (LC) is given by:
$\text{LC} = 1\text{ MSD} - 1\text{ VSD}$
$\text{LC} = 1\text{ MSD} - \dfrac{19}{20}\text{ MSD} = \dfrac{1}{20}\text{ MSD}$

We are given that $\text{LC} = 0.005\text{ cm}$. Therefore:
$\dfrac{1}{20}\text{ MSD} = 0.005\text{ cm}$
$1\text{ MSD} = 20 \times 0.005\text{ cm} = 0.1\text{ cm}$

The number of main scale divisions in $1\text{ cm}$ is:
$\text{Number of divisions} = \dfrac{1\text{ cm}}{0.1\text{ cm}} = 10$

Answer: $10$