The pitch of a screw gauge is $0.5\text{mm}$ and there are $50$ divisions on its circular scale. The screw gauge has the least count of

The correct option is C $0.01\text{mm}$
Least count of a screw gauge is given as,
$\text{LC}=\frac{\text{Pitch}}{\text{Number of divisions on circular scale}}$

$=\frac{0.5}{50}$

$=0.01\text{mm}$