The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line. The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is:

Engineering

Physics

Errors

Question

The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line.

The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is:

5.950 mm

5.740 mm

5.725 mm

5.755 mm

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$Zero error = + 3 \times \frac{0 .5 mm}{100} = 0 .015 mm$

$MSR = 5 .5 + 48 \times \frac{0 .5}{100}$

= 5.74 mm.

$∴$ Thickness = 5.74 – 0.015 = 5.725 mm