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The pitch of an error free screw gauge is 1 mm and there are 100 divisions on the circular scale. While measuring the diameter of a thick wire, the pitch scale reads 1 mm and $63^{\text {rd }}$ division on the circular scale coincides with the reference line. The diameter of the wire is:
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The correct answer is:
0.163 cm
Least count of screw gauge $=\frac{\text { Pitch }}{\text { No. of divisions on circular scale }}$
$\Rightarrow \quad$ Least count $=\frac{1}{100}=0.01 \mathrm{~mm}$
Final reading $=$ MSR $+\mathrm{CSR} \times$ L.C.
$=1 \mathrm{~mm}+(63)(0.01) \mathrm{mm}$
$=1.63 \mathrm{~mm}$
$=0.163 \mathrm{~cm}$
$\Rightarrow \quad$ Least count $=\frac{1}{100}=0.01 \mathrm{~mm}$
Final reading $=$ MSR $+\mathrm{CSR} \times$ L.C.
$=1 \mathrm{~mm}+(63)(0.01) \mathrm{mm}$
$=1.63 \mathrm{~mm}$
$=0.163 \mathrm{~cm}$
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