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 :

• A. 5.755 m
• B. 5.725 mm
• C. 5.740 m
• D. 5.950 mm

Answer 1

Answer: (B)

5.725 mm

$LC=\frac{Pitch}{No.ofdivision}$
$LC=0.5\times {10}^{-2}mm$
+ve error = $3\times 0.5\times {10}^{-2}mm$
$=1.5\times {10}^{-2}mm=0.015mm$
Reading = MSR + CSR - (+ve error)
$=5.5mm+(48\times 0.5\times {10}^{-2})-0.015$
$=5.5+0.24-0.015=5.725mm$