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Thedistancebetween twoconsecutivethreadson thescrewof ascrewgaugeis 0.5mm. Thenumberofdivisionson thecircular scaleis 100. a wireis placedbetween of thestudsof the screw gauge.Find thediameter ofthe wireif thepitchscaleshows 14th division and 40th circulardivision coincideswiththe baseline. Thegivenapparatus wasdetected to havenegative zeroerror. The 90th division on the circular scale of coincides withreferenceline, whenthe studsare incontact. |
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Answer» SOLUTION :1 MSD=1 mm `" Pitc " = (5 MSD)/(5) =1 MSD =1MM` Leastcount `= (P)/(N)` `N=(1 mm)/(1 MU m) =100` N=500 , P =(N) `xx` (L.C.) =500 `xx 1 mu m = 0.5` mm PITCH =0.5 mm |
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