1.

The pitch of a screw gauge is `0.55mm` and there are 100 divisions on its 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. Then the diameter of the wire isA. `4.05mm`B. `4.405mm`C. `3.05mm`D. `1.25mm`

Answer» `Deltal = 0.5mm`
`N = 100 divisions`
zero correction = 2 divisions Reading =Measured value + zero correction .
`=(8 xx 0.5)mm + (83 -2) xx (0.5)/(100)`
`4mm + 81 xx (0.5)/(100)mm`
`=4.405mm` .


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