Question

In a screw gauge, there are 100 divisions on the circular scale and the main scale moves by $05 \,mm$ 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

The correct answer is 2.2
Least count = No. of circular divisions Pitch ​
=1000.5mm​
Least count =5×10−3mm
Positive Error =MSR+CSR(LC)
=0mm+6(5×10−3mm)
Reading of Diameter = MSR +CSR(LC)−
Positive zero error
=4×0.5mm+(46(5×10−3))−6(5×10−3)mm
=2mm+40×5×10−3mm=2.2mm