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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 ....... $mm$
$3.38$
$3.32$
$3.73$
$3.67$
Solution
Least count of screw gauge $= \frac{{0.5}}{{50}}mm = 0.01mm$
$\therefore$ Reading $ = \,\left[ {Main\,scale\,\,reading + circular\,scale\,reading \times L.C} \right] – \left( {zero\,error} \right)$
$ = \left[ {3 + 35 \times 0.01} \right] – \left( { – 0.03} \right) = 3.38\,mm$
