JEE MAIN - Physics (2008 - No. 31)
Two full turns of the circular scale of a screw gauge cover a distance of 1 mm on its main scale. The
total number of divisions on the circular scale is 50. Further, it is found that the screw gauge has a
zero error of − 0.03 mm while measuring the diameter of a thin wire, a student notes the main scale
reading of 3 mm and the number of circular scale divisions in line with the main scale as 35. The
diameter of the wire is
3.32 mm
3.73 mm
3.67 mm
3.38 mm
Explanation
Least count of screw gauge = $${{0.5} \over {50}}mm$$ = 0.01mm
Main scale reading = 3 mm
Vernier scale reading = 35
$$\therefore $$ Reading = [Main scale reading + circular scale reading $$\times$$ L.C] - (zero error)
= [3 + 35 $$\times$$ 0.01] - (-0.03) = 3.38 mm
Main scale reading = 3 mm
Vernier scale reading = 35
$$\therefore $$ Reading = [Main scale reading + circular scale reading $$\times$$ L.C] - (zero error)
= [3 + 35 $$\times$$ 0.01] - (-0.03) = 3.38 mm
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