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Q.

The pitch of a screw gauge is 0.5 mm and there are 50 divisions on its circular scale and one main scale division = 0.5 mm. Before starting the measurement it is found that when and zero of the main scale is hardly visible jaws of the screw gauge are brought in contact, the zero of the circular scale lies 4 division above the reference line. When a metallic wire is placed between the jaws, five main scale divisions are clearly visible and
18th division on the circular scale coincides with the reference line. The diameter of the wire is K100milli meter. Then K is

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answer is 272.

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Detailed Solution

Zero error =+4×0.01=+0.04mm
Measured diameter =MSR+ L.C. ×CSR
=(5×0.5)mm+(0.01mm)×18=2.68mm
 Actual diameter =2.68+(0.04)=2.64mm

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