The least count of a screw gauge is 0.005mm and it has 100 equal divisions on its head scale. Then the distance between two consecutive threads on its screw is
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Answer:
For d
1
:
d
1
=Pitch=Distance between consecutive threads
=(least count) × (total number of division on head scale)
=(0.005 mm) (200)
=1 mm
For d
2
:
d
2
= Vernier constant
=1mm−(
100
99
)mm
=0.01 mm
For d
3
:
p=1mm
N=100
Least count =
N
p
=
100
1mm
=0.01mm
Main scale reading =MSR=2×(1mm)=2mm
Circular scale reading =CSR=67×(0.01)=0.67mm
d 3
=MSR+CSR
=2+0.67
=2.67mm
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