The given 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 $$1$$ mm when the circular scale makes $$2$$ complete revolutions, the reading of the instrument in the figure is $$____mm$$

Question

Infinity Learn · Accepted Answer

Pitch of the screw $$=1$$ $$mm2=0.5$$ mm​ Least count $$=0.550=0.01$$ mm​ ∴Reading $$=$$ Linear scale reading $$+$$ (coinciding$$ circular scale × least $$count) ​                    $$=3.0$$ $$+32$$×$$0.01$$​                    $$=3.0+0.32$$​                        $$=3.32$$ mm​

Measurement

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The given 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 mm when the circular scale makes 2 complete revolutions, the reading of the instrument in the figure is ____mm

$\begin{array}{l} Pitch of the screw = \frac{1 m m}{2} = 0.5 m m \\ Least count = \frac{0.5}{50} = 0.01 m m \\ ∴ Reading = Linear scale reading + (coinciding circular scale \times least count) \\ = 3.0 + (32 \times 0.01) \\ = 3.0 + 0.32 \\ = 3.32 m m \end{array}$

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Measurement

Remember concepts with our Masterclasses.

The given 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 mm when the circular scale makes 2 complete revolutions, the reading of the instrument in the figure is ____mm

Pitch of the screw =1 mm2=0.5 mm​ Least count =0.550=0.01 mm​ ∴Reading = Linear scale reading + (coinciding circular scale × least count) ​ =3.0 +32×0.01​ =3.0+0.32​ =3.32 mm​

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$\begin{array}{l} Pitch of the screw = \frac{1 m m}{2} = 0.5 m m \\ Least count = \frac{0.5}{50} = 0.01 m m \\ ∴ Reading = Linear scale reading + (coinciding circular scale \times least count) \\ = 3.0 + (32 \times 0.01) \\ = 3.0 + 0.32 \\ = 3.32 m m \end{array}$