Question 201: Units , Dimensions and Error Analysis

Question

In a screw gauge, $$5$$ complete rotations of the screw cause it to move a linear distance of $$0.25$$ $$cm.$$ There are $$100$$ circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of $$4$$ main scale divisions and $$30$$ circular scale divisions. Assuming negligible zero error, the thickness of the wire is :

Accepted Answer

5 complete rotations = 0.25 cms
So, 1 complete rotation of screw = 0.05 cm
$$	herefore\,\,\,\,$$  1 main scale division = 0.05 cm
1 circular scale = $${{0.05} \over {100}}$$ = 5 $$ 	imes $$ 10
$$-$$4
cm
Thickness of a wire
= 4 main scale and 30 circular scale divisions
= 4 $$ 	imes $$ 0.05 + 30 $$ 	imes $$ 5 $$ 	imes $$ 10
$$-$$4
= 0.2150 cm.

Answer

$$0.4300$$ $$cm$$

Answer

$$0.2150$$ $$cm$$

Answer

$$0.3150$$ $$cm$$

Answer

$$0.0430$$ $$cm$$

Units , Dimensions and Error Analysis

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