The pitch of a screw gauge is $$1$$ mm and there are $$100$$ divisions on circular scale. When faces A and B are just touching each other without putting anything between the studs $$32$$ nd division of the circular scale coincides with the reference line. When a glass plate is placed between the studs, the linear scale reads $$4$$ divisions and the circular scale reads $$16$$ divisions. Find the thickness of the glass plate. Zero of linear scale is not hidden from circular scale when A and B touches each other.

Question

The pitch of a screw gauge is $$1$$ mm and there are $$100$$ divisions on circular scale. When faces A and B are just touching each other without putting anything between the studs $$32$$ nd division of the circular scale coincides with the reference line. When a glass plate is placed between the studs, the linear scale reads $$4$$ divisions and the circular scale reads $$16$$ divisions. Find the thickness of the glass plate. Zero of linear scale is not hidden from circular scale when A and B touches each other.

Infinity Learn · Accepted Answer

Least count (L$$ $$C) $$=$$ Pitch  Number of divisions on circular scale $$=1100mm=$$ $$0.01$$ mmAs zero is not hidden from circular scale when A and B touches each other. Hence, the screw gauge has positive error.$$e=+n(LC)=32$$×$$0.01=0.32$$ mmLinear scale reading $$=4$$×$$(1$$ $$mm)=4$$ mmCircular scale reading $$=16$$×$$(0.01$$ $$mm)=0.16$$ mmMeasured reading $$=(4+0.16)mm=4.16$$ mmAbsolute reading $$=$$ Measured reading $$-e=(4.16-0.32)mm=3.84$$ mmTherefore, thickness of the glass plate is $$3.84$$ mm.

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