Q. 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 :

Solution:

LC = $\frac{\text{Pitch}}{\text{NO} \text{of} \text{division}} $
$ LC = 0.5 \times10^{-2} mm $
+ve error = $3 \times 0.5 \times 10^{-2} mm$
$= 1.5 \times 10^{-2} \; mm = 0.015 mm$
Reading = MSR + CSR - (+ve error)
$= 5.5 mm + (48 \times 0.5 \times 10^{-2}) - 0.015$
= 5.5 + 0.24 - 0.015 = 5.725 mm

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