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The pitch of a screw gauge is 1 mm and there are \({100}\text{divisions}\) on circular scale. When faces \({A}\) and \({B}\) are just touching each without putting anything between the studs 32nd divisions of the circular scale (below its Zero) coincides with the reference line. When a glass plate is placed between the studs, the linear scale reads 4 divisions and the circular reads 16 divisions. Find the thickness of the glass plate. Zero of linnear scale is not hidden from circular scale when A and B touches each other.

Answer

Least count (LC) \(=\frac{{\text{Pitch}}}{{\text{Number of divisions on circular scale}}}=\frac{{{1}}}{{{100}}}{m}{m}\)
=\({0.01}{m}{m}\)
As zero is not hidden from circular scale when A and B touches each other. Hence the screw gauge has positive error.
\({e}=+{n}{\left({L}{C}\right)}={32}\times{0.01}={0.32}{m}{m}\)
Linear scale reading =\({4}\times{\left({1}{m}{m}\right)}={4}{m}{m}\)
Circular scale reading =\({16}\times{\left({0.01}{m}{m}\right)}={0.16}{m}{m}\)
\(\therefore\) Measured reading =\({\left({4}+{0.16}\right)}{m}{m}={4.16}{m}{m}\)
\(\therefore\) Absolute reading = Measured reading - e
\(={\left({4.16}-{0.32}\right)}{m}{m}={3.84}{m}{m}\)
Therefore, thickness of the glass plate is \({3.84}{m}{m}\).
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