Question

a screw gauge has 50 divisions on its circular scale the pitch of the screw gauge in 0.5 mm what is its least count​

Answer 1

Answer:

0.5 mm would be the least count.

Answer 2

Answer:

mate your answer,

Explanation:

Pitch of the circular scale  P=0.5 mm

Number of divisions on circular scale  N=50

So least count of screw gauge   L.C =  P/N

​ 0.5/50 mm=

0.01 mm

Now, main scale reading is given as  M.S.R =2 mm

Marking on circular scale  n=25

Thus, diameter of sphere  d=M.S.R+n×L.C

⟹ d=2 mm+25×0.01 mm=2.25 mm.

Hope it helps.