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mihirthemega:
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Given,
distance travelled on main scale = 20 mm
No. of rotations = 10
So, Pitch of the screw gauge = 20 mm/ 10 = 2 mm
No. of divisions on circular scale = 50
So, Least Count (L.C.) of the screw gauge = 2 mm/50 = 1/25 mm = 4/100 mm = 0.04 mm
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