Match List-I with List-II regarding the least count of various...

For a screw gauge, Least Count (LC) = $\dfrac{\text{Pitch}}{\text{Number of circular divisions}}$

(A) $\text{LC} = \dfrac{1\text{ mm}}{100} = 0.01\text{ mm}$. This matches (III).

(B) $\text{LC} = \dfrac{0.5\text{ mm}}{100} = 0.005\text{ mm}$. This matches (IV).

For a Vernier caliper, Least Count (LC) = $1\text{ MSD} - 1\text{ VSD}$

(C) $10\text{ VSD} = 9\text{ MSD} \Rightarrow 1\text{ VSD} = 0.9\text{ MSD}$.
$\text{LC} = 1\text{ MSD} - 0.9\text{ MSD} = 0.1\text{ MSD}$. Since $1\text{ MSD} = 1\text{ mm}$, $\text{LC} = 0.1\text{ mm}$. This matches (II).

(D) $50\text{ VSD} = 49\text{ MSD} \Rightarrow 1\text{ VSD} = \dfrac{49}{50}\text{ MSD}$.
$\text{LC} = 1\text{ MSD} - \dfrac{49}{50}\text{ MSD} = \dfrac{1}{50}\text{ MSD} = 0.02\text{ mm}$. This matches (I).

Therefore, the correct matching is (A)-(III), (B)-(IV), (C)-(II), (D)-(I).

Answer: (A)-(III), (B)-(IV), (C)-(II), (D)-(I)