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: array|c|c|c| & 2c|Readings \ 2-3 & LS (mm) & CS \ Wire-1 & 0.5 & 42 \ Wire-2 & 1.5 & 95 \ array
Answer
1915
Solution
Key idea: least count = pitchCS divisions= 0.5100=0.005 mm; reading =LS+CS ; the known Wire-1 gives the zero error.
Wire-1 reading =0.5+42 0.005=0.710 mm; true value 0.650 mm, so zero error =+0.060 mm.
Wire-2 reading =1.5+95 0.005=1.975 mm.
d=1.975-0.060=1.915 mm=1915 .
Answer: 1915