Pitch of a screw gauge is $$1$$ mm and there are $$100$$ divisions on it’s circular scale. When nothing is kept between the studs, the zero of the main scale is not visible and $$14th$$ division of circular scale coincides with the horizontal mark on main scale. When circular scale is rotated by one complete rotation, zero mark on main scale is still not visible, but during the second rotation zero mark becomes just visible and $$10th$$ division of circular scale is coinciding with horizontal mark on main scale. When a glass plate is placed between the studs, the circular scale lies between $$8th$$ and $$9th$$ division of main scale and $$4th$$ division of circular scale coincides with horizontal mark on main scale. Main scale division is equal to $$1mm$$. Then

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Infinity Learn · Accepted Answer

pitch of the $$screw=distance$$ moved by the screw on pitch scale No. of complete rotations of $$screw=1mm1=1mmLC=pitch$$ of the screwNo. of Circularscale $$divisions=1mm100=0.01mmIn$$ $$1st$$ full rotation, circular scale is moved $$by100$$ divisions. In $$2nd$$ rotation, circular scale moves by $$96$$ more divisions.Total length of main scale for $$196$$ division $$=$$ $$196$$×$$0.01mm=1.96mmZero$$ error  $$=$$ $$-1.96$$ mmThickness $$=$$[$$MSR+CSR(LC)$$ ]$$+$$ $$correction=8+4(0.01)+1.96=8.04$$ $$+$$ $$1.96$$ $$=$$ $$10.00$$ mm

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