A screw gauge has as many circular scale divisions as there are threads in 5 cm length of the screw . If the pitch of the screw is 0.5 mm then what is its least count ?

Options

A. 0.001 mm
B. 0.001 cm
C. 0.0005 mm
D. 0.0005 cm

Explanation

Let the n rotations advance the screw by 5 cm
`:. " Pitch "=(5)/(n) cm`
But pitch = 0.5 mm = 0.05 cm
`:. (5)/(n)= 0.05`
` n=10`
Thus there are 100 threads in 5 cm of the screw i.e., 100 circular scale divisions
`:. L.C. =(0.05)/(100) =0.0005 cm`