The accompanying diagram represents a screw gauge. The circular scale is divided into $50$ divisions and the linear scale is divided into millimeters. If the screw advances by $1 m m$ when the circular scale makes $2$ complete revolutions, the least count of the instrument and the reading of the instrument in figure are:

0.01 m m, 3.82 m m

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0.01 m m, 4.82 m m

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0.02 m m, 3.82 m m

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0.05 m m, 3.82 m m

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Solution

The correct option is D $0.01 m m, 3.82 m m$
Pitch of the screw gauge, $p = \frac{1 m m}{2} = 0.5 m m$
Least count, $L c = p / n = 0.5 / 50 = 0.01 m m$
From diagram, linear scale reading, $L S R = 3 m m$ and circular scale reading, $C S R = 50 + 32 = 82$
Total reading, $R = L S R + (C S R \times L c) = 3 + (82 \times 0.01) = 3.82 m m$

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The correct option is D 0.01mm,3.82mmPitch of the screw gauge, p=1mm2=0.5mmLeast count, Lc=p/n=0.5/50=0.01mmFrom diagram, linear scale reading, LSR=3mm and circular scale reading, CSR=50+32=82Total reading, R=LSR+(CSR×Lc)=3+(82×0.01)=3.82mm