Question

A screw gauge has 100 divisions on circular scale and pitch is 0.5 mm. When we fix a wire between its jaws, it reads 2 divisions on the main scale and 48 divisions on a circular scale. If the screw gauge has no zero error, the radius of the wire is:-

Answer 1

Least count of screw gauge  = pitch /no of divisions

                                             =    .5 /100      

                                             = .005 mm

Length of wire = msr +csr ×lc

                          = 2 + 48 × .005

                           = 2 + .24

                            = 2.24 mm

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Q.1. A screw gauge is having 100 divisions on its circular scale and each main scale division is 1 mm. Least count of the screw gauge is

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