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Question
Figure shows a screw gauge in which circular scale has 200 divisions. Calculate the least count and radius of the wire.
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Solution
No. of circular scale divisions = 200
Pitch = 1 mm
Least count (L.C.) = `"Pitch"/"No. of circular divisions"`
= `(1 "mm")/200`
= 0.005 mm
L.C. = 0.0005 cm
Main scale reading = 5 mm = 0.5 cm
(C.S.D.) circular scale reading = 34 divisions
Observed diameter of wire = Main scale reading + L.C. × C.S.D.
= 0.5 + 0.0005 × 34
= 0.5 + 0.0170
= 0.5170 cm
Redius of wire = `"diameter"/2`
= `0.5170/2`
= 0.2585 cm
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