Q122 / 200 · NEET 2024 (Re-NEET) · Physics

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:

  • A
    1.63 cm
  • Correct answer
    B
    0.163 cm
  • C
    0.163 m
  • D
    1.63 m

Solution

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. $=1 \mathrm{~mm}+(63)(0.01) \mathrm{mm}$ $=1.63 \mathrm{~mm}$ $=0.163 \mathrm{~cm}$

Hence, the correct answer is option: 'B'.

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