The least count of a vernier caliper is 0.005 cm . It is given that...

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

Given that $20 \text{ VSD} = 19 \text{ MSD}$
$\Rightarrow 1 \text{ VSD} = \dfrac{19}{20} \text{ MSD}$

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 $\text{LC} = 0.005 \text{ cm}$.
$\Rightarrow \dfrac{1}{20} \text{ MSD} = 0.005 \text{ cm}$
$\Rightarrow 1 \text{ MSD} = 20 \times 0.005 \text{ cm} = 0.1 \text{ cm}$

The length of one main scale division is $0.1 \text{ cm}$.
Therefore, 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: $10$