Please solve question no.15

Dear Student

Ans15.
Given
Pitch of screw gauge, P = 0.5 mm
Least count, LC = 0.001 mm
We Know that
$LC=\frac{P}{N}$
Where
LC = Least count of screw gauge
P = Pitch of the screw gauge
N = Number of divisions on circular scale
Therefore,
Number of divisions on circular scale, $N=\frac{P}{LC}=\frac{0.5 mm}{0.001 mm}=500$

Regards