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प्रश्न
A screw gauge has 50 divisions on its circular scale and its screw moves by 1 mm on turning it by two revolutions. When the flat end of the screw is in contact with the stud, the zero of the circular scale lies below the base line and 4th division of the circular scale is in line with the base line. Find
(i) The pitch,
(ii) The least count and
(iii) The zero error of the screw gauge
उत्तर
No. of divisions on the circular scale = 50
(i) Pitch = Distance moved ahead in one revolution
= 1 mm/2 = 0.5 mm
(ii) L.C. = Pitch/No. of divisions on the circular head
= (0.5/50) mm
= 0.01 mm
(iii) Because the zero of the circular scale lies below the base line, when the flat end of the screw is in contact with the stud, the error is positive.
No. of circular division coinciding with m.s.d. = 4
Zero error = + (4 × L.C.)
= + (4 × 0.01) mm
= + 0.04 mm
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