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Question

The pitch of a screw gauge is 1 mm and there are 100 divisions on circular scale. When the faces A and B are just touching each without putting anything between the studs, the 32nd division of the circular scale coincides with the reference line. The error in measurement due to screw gauge will be (zero of linear scale is not hidden from circular scale when A and B touches)


A
0.04 mm
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B
+0.32 mm
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C
+0.16 mm
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D
+0.04 mm
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Solution

The correct option is B +0.32 mm
Least count (LC) = pitch/ number of divisions on circular scale
LC=1100 mm=0.01 mm
When A and B touches each other, zero is not hidden from the circular scale. Thus, the screw gauge is having a positive error.
e=n×LC
e=+32×0.01=+0.32 mm
Thus correct reading will be = Measured - Positive error

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