• # question_answer 41) The circular head of a screw gauge is divided into 200 divisions and move $1\text{ }mm$ ahead in one revolution. If the same instrument has a zero error of- $0.05\text{ }mm$ and the reading on the main scale in measuring diameter of a wire is $6\text{ }mm$ and that on circular scale is 45. The diameter of the wire is A)  $6.275\,mm$               B)  $6.375\,mm$C)  $5.75\,mm$                D)  $5.50\,mm$

Pitch = 1 mm                                        Number of divisions on circular scale = 200 $L.C=\frac{Pitch}{Number\,ofdivisions\,on\,circular\,scale}$ $=\frac{1\,mm}{200}=0.005\,mm=0.0005\,cm$ Diameter of the wire = (Main scale reading + Circular scale reading $\times$ L.C.) - zero error $=6\,mm+45\times 0.05-(-0.05)$ $=6\,mm+0.225mm+0.05mm=6.275\,mm$