In an experiment to find out the diameter of wire using screw gauge, the following observation were noted. \((a)\) Screw moves \(0.5\,mm\) on main scale in one complete rotation \((b)\) Total divisions on circular scale \(=50\) \((c)\) Main scale reading is \(2.5\,mm\) \((d)\) \(45^{\text {th }}\) division of circular scale is in the pitch line \((e)\) Instrument has \(0.03 \;mm\) negative error Then the diameter of wire is \(...........\,mm\)

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Answer & Solution

Correct Answer

(C) \(2.98\)

Step-by-step Solution

Detailed explanation

\(MSR =2.5\,mm\)

\(CSR =45 \times \frac{0.5}{50}\,mm\)

\(=0.45\,mm\)

Diameter reading \(= MSR + CSR -\) zero error

\(=2.5+0.45-(-0.03)\)

\(=2.98\,mm\)

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