Question 193: Units , Dimensions and Error Analysis

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 :

Accepted Answer

We know,
Least count (LC) = $${{Pitch} \over {no.\,of\,divisions}}$$
$$ 	herefore $$  LC = $${{0.5} \over {100}}$$
= 0.5 $$ 	imes $$ 10
$$-$$2
mm
Reading = MSR + CSR $$-$$ positive error
Given, Main scale reading (MSR) = 5.5 mm
Circular scale reading (CSR)
= 48 $$ 	imes $$ 0.5 $$ 	imes $$ 10
$$-$$2
mm
= 0.24
As zero of its circular scale lines 3 division below the mean line, it means error is position error.
$$ 	herefore $$  positive error
= 3 $$ 	imes $$ 0.5 $$ 	imes $$ 10
$$-$$2
mm
= 0.015 mm
$$ 	herefore $$  Reading = 5.5 + 0.24 $$-$$ 0.015
= 5.725 mm

Answer

5.755 mm

Answer

5.950 mm

Answer

5.725 mm

Answer

5.740 mm

Units , Dimensions and Error Analysis

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