Question

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 _____$\times 10^{-2} mm$.

Answer 1

Least count $=\frac{\text { Pitch }}{\text { No. of circular divisions }}$
$=\frac{0.5 mm }{100}$
Least count $=5 \times 10^{-3} mm$
Positive Error $= MSR + CSR ( LC )$
$=0 mm +6\left(5 \times 10^{-3} mm \right)$
Reading of Diameter $=$ MSR $+\operatorname{CSR}( LC )-$
Positive zero error
$=4 \times 0.5 \,mm +\left(46\left(5 \times 10^{-3}\right)\right)-6\left(5 \times 10^{-3}\right) \,mm$
$=2 \,mm +40 \times 5 \times 10^{-3} mm =2.2 \,mm$