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In a screw gauge, there are $100$ divisions on the circular scale and the main scale moves by $0.5\,mm$ on a complete rotation of the circular scale. The zero of circular scale lies $6$ divisions below the line of graduation when two studs are brought in contact with each other. When a wire is placed between the studs, $4$ linear scale divisions are clearly visible while $46^{\text {th }}$ division the circular scale coincide with the reference line. The diameter of the wire is $...........\times 10^{-2}\,mm$
$23$
$20$
$21$
$22$
Solution
$\text { Least count }=\frac{\text { Pitch }}{\text { No. of circular divisions }}$
$=\frac{0.5\,mm }{100}$
$\text { Least count }=5 \times 10^{-3}\,mm$
$\text { Positive Error }= MSR + CSR ( LC )$
$=0\,mm +6\left(5 \times 10^{-3}\,mm \right)$
$\text { Reading of Diameter }= MSR + CSR ( LC )-$
$\text { Positive zero error }$
$=4 \times 0.5\,mm +\left(46\left(5 \times 10^{-3}\right)\right)-6\left(5 \times 10^{-3}\right)\,mm$
$=2\,mm +40 \times 5 \times 10^{-3}\,mm =2.2\,mm$
