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  4. The pitch of a screw gauge is 0.5 mm and there are 100 divisions on it circular scale. The instrument reads +2 divisions when nothing is put in between its jaws. In measuring the diameter of a wire, there are 8 divisions on the main scale and 83rd division coincides with the reference line. Then the diameter of the wire is

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Q. The pitch of a screw gauge is $0.5\, mm$ and there are $100$ divisions on it circular scale. The instrument reads $+2$ divisions when nothing is put in between its jaws. In measuring the diameter of a wire, there are $8$ divisions on the main scale and $83^{rd}$ division coincides with the reference line. Then the diameter of the wire is

Physical World, Units and Measurements

Solution:

$\Delta l= 0.5 \,mm$
$N= 100$ divisions
Zero correction $= 2$ divisions
Reading = Measured value + Zero correction
$= \left(8\times0.5\right)mm + \left(83-2\right)\times\frac{0.5}{100}$
$=4mm+81\times\frac{0.5}{100}$
$=4.405\,mm$