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

Correct Answer: 2.2

The correct answer is 22.
Least count $=\frac{\text{Pitch}}{\text{No. of circular divisions}}$
$=\frac{0.5mm}{100}$
Least count $=5\times 10^{-3}mm$
Positive Error $=MSR+CSR(LC)$
$=0mm+6\left(5\times 10^{-3}mm\right)$
Reading of Diameter $=$ MSR $+\mathrm{CSR}(LC)-$
Positive zero error
$=4\times 0.5mm+\left(46\left(5\times 10^{-3}\right)\right)-6\left(5\times 10^{-3}\right)mm$
$=2mm+40\times 5\times 10^{-3}mm=2.2mm$