The pitch of the screw gauge is \(1\, mm\) and there are \(100\) divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lies \(8\) divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while \(72^{\text {nd }}\) division on circular scale coincides with the reference line. The radius of the wire is.........\(mm\)

Pick an option to check your answer.

Verified Solution

Answer & Solution

Correct Answer

(B) \(0.82\)

Detailed explanation

Least count \(=\frac{1 mm }{100}=0.01 mm\)

zero error \(=+8 \times LC =+0.08 mm\)

True reading (Diameter)

\(=(1 mm +72 \times LC )-(\) Zero error \()\)

\(=(1 mm +72 \times 0.01 mm )-0.08 mm\)

\(=1.72 mm -0.08 mm\)

\(=1.64 mm\)

therefore, radius \(=\frac{1.64}{2}=0.82 mm\)

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