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A screw gauge with a pitch of `0.5mm` and a circular scale with `50` divisions is used to measure the thicknes of a thin sheet of Aluminium. Before starting the measurement, it is found that wen the jaws of the screw gauge are brought in cintact, the `45^(th)` division coincide with the main scale line and the zero of the main scale is barely visible. what is the thickness of the sheet if the main scale readind is `0.5 mm` and the `25th` division coincide with the main scale line?A. `0.75mm`B. `0.80mm`C. `0.70mm`D. `0.50mm` |
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Answer» Correct Answer - B Screw gauge has negative zero error least count of screw gauge `LC=("Pitch")/("Number of divisions on circular scale")` `LC=(0.5mm)/(50)=0.01mm` zero error`=(45-50)xx0.01mm=-0.05mm` Thickness of sheets=Main Scale reading `+`(cuircular scale reading `xx1C`)`-`zero error `=0.5+(25xx0.01)-(-0.05)` `=0.5+0.30=0.80mm` |
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