Questions
The pitch of a screw gauge is 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.
detailed solution
Correct option is D
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.Talk to our academic expert!
Similar Questions
The distance moved by the screw of a screw gauge is 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
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