The distance moved by the screw of a screw gauge is $$2$$ mm in four rotations and there are $$50$$ divisions on its cap. When nothing is put between its jaws, $$20th$$ division of circular scale coincides with reference line and zero of linear scale is hidden from circular scale when two jaws touch each other or zero of circular scale is lying above the reference line. When plate is placed between the jaws, main scale reads $$2$$ divisions and circular scale reads $$20$$ divisions. Thickness of plate is

Question

The distance moved by the screw of a screw gauge is $$2$$ mm in four rotations and there are $$50$$ divisions on its cap. When nothing is put between its jaws, $$20th$$ division of circular scale coincides with reference line and zero of linear scale is hidden from circular scale when two jaws touch each other or zero of circular scale is lying above the reference line. When plate is placed between the jaws, main scale reads $$2$$ divisions and circular scale reads $$20$$ divisions. Thickness of plate is

Infinity Learn · Accepted Answer

Distance moved in one rotation $$=$$ $$0.5$$ mmLeast count, LC $$=$$ $$0.5$$ $$mm50$$ divisions $$=0.01$$ mmScrew gauge has negative zero error.This error is $$(50-20)0.01$$ mm or $$(30)(0.01)mm$$.Thickness of $$plate=(2$$×$$0.5$$ $$mm)+(30+20)(0.01$$ $$mm)=$$ $$1.5$$  mm

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