First slide
Measurement
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

Moderate
Solution

Distance moved in one rotation = 0.5 mm

Least count, LC = 0.5 mm50 divisions =0.01 mm

Screw 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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