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A screw gauge have 250 equal divisions marked along the periphery of circular scale. If one full rotation gives advancement of 0.625 cm, the least count of the arrangement is

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2.5×10−4cm

2.5×10−2cm

2.5×10−3cm

None of these

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detailed solution

Correct option is C

Least count =Pitch of the screwno.of divisions on circular scale =0.625250cm =0.0025 cm =2.5×10-3 cm

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