Question

The circular scale of a screw gauge has 50 divisions. It's spindle moves by 2 mm on sleeve. when given four complete rotations calculate

pitch

least count.​

Answer 1

Answer:

2.25

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 =NP​=500.5​ 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