Two thin wires, Wire-1 of diameter 0.650 mm and Wire-2 of unknown diameter d are given. To obtain the value of d , the diameters of the two wires are measured with a screw gauge. The screw gauge has a pitch of 0.5 mm and there are 100 divisions on the circular scale (CS). The smallest division on the linear scale (LS) is 0.5 mm. The table shows the readings of LS and CS for the measurements. The value of d (in m) is:

Step-by-step solution

The least count (LC) of the screw gauge is given by: LC = Pitch Number of divisions on CS = 0.5 100 = 0.005 mm For Wire-1, the measured reading is: Reading ₁ = LS + CS LC Reading ₁ = 0.5 + 42 0.005 = 0.5 + 0.210 = 0.710 mm The actual diameter of Wire-1 is 0.650 mm . The zero error of the screw gauge is: Zero Error = Measured Reading - Actual Value = 0.710 - 0.650 = +0.060 mm For Wire-2, the measured reading is: Reading ₂ = LS + CS LC Reading ₂ = 1.5 + 95 0.005 = 1.5 + 0.475 = 1.975 mm The actual diameter d of Wire-