JEE MAIN - Physics (2018 - 15th April Morning Slot - No. 7)
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 :
$$0.4300$$ $$cm$$
$$0.2150$$ $$cm$$
$$0.3150$$ $$cm$$
$$0.0430$$ $$cm$$
Explanation
5 complete rotations = 0.25 cms
So, 1 complete rotation of screw = 0.05 cm
$$\therefore\,\,\,\,$$ 1 main scale division = 0.05 cm
1 circular scale = $${{0.05} \over {100}}$$ = 5 $$ \times $$ 10$$-$$4 cm
Thickness of a wire
= 4 main scale and 30 circular scale divisions
= 4 $$ \times $$ 0.05 + 30 $$ \times $$ 5 $$ \times $$ 10 $$-$$4
= 0.2150 cm.
So, 1 complete rotation of screw = 0.05 cm
$$\therefore\,\,\,\,$$ 1 main scale division = 0.05 cm
1 circular scale = $${{0.05} \over {100}}$$ = 5 $$ \times $$ 10$$-$$4 cm
Thickness of a wire
= 4 main scale and 30 circular scale divisions
= 4 $$ \times $$ 0.05 + 30 $$ \times $$ 5 $$ \times $$ 10 $$-$$4
= 0.2150 cm.
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