Question

The pitch of a screw gauge having 50 divisions on its circular scale is 1 mm When the two jaws of the screw gauge are in contact with each other, the zero of the circular scale lies 6 divisions below the line of gradution. when a wire is placed between the jaws, 3 linear scale divisions are clearly visible while 31 division on the circular scale coincides with the reference line. Find diameter of the wire.

Answer 1

Given :

Circular scale divisions = 50

Pitch = 1mm

n=6

Linear scale reading = 3

To find :

The diameter of the wire

Solution :

• Least count of screw guage = (pitch/No of circular scale divisions)

       L.C= 1mm/50

       L.C=0.02mm

• The screw guage has positive error

     Error = n×L.C

              = 6× 0.02mm

     Error =0.12mm

• Circular scale reading = 31×0.02mm =0.62mm

• Diameter of wire = Linear scale reading + circular scale reading

                            =3 + 0.62 = 3.62mm

• Actual diameter of wire = 3.62 - error

                                       =3.62 - 0.12

                                       =3.50mm

     The diameter of the wire is 3.50mm

Answer 2

Explanation:

given pitch equals to 1 mm

CSD = 50

therefore least count equals to pitch by CSD

equals to 1 /50=0.02mm

1MSR= 3mm

1 CSR = 31

error(e)= 6×L.C

= 6×0.02

= 0.12mm