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The pitch of a screw gauge is 1 mm and t...

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 without putting anything between the studs 32nd divisions of the circular scale (below its Zero) coincides with the reference line. When a glass plate is placed between the studs, the linear scale reads 4 divisions and the circular reads 16 divisions. Find the thickness of the glass plate. Zero of linnear scale is not hidden from circular scale when A and B touches each other.

Text Solution

Verified by Experts

Least count (LC) `=("Pitch")/("Number of divisions on circular scale")=(1)/(100) mm`
=`0.01 mm`
As zero is not hidden from circular scale when A and B touches each other. Hence the screw gauge has positive error.
`e = +n(LC)=32xx0.01=0.32mm`
Linear scale reading =`4xx (1 mm)=4mm`
Circular scale reading =`16xx (0.01mm) =0.16mm`
`:.` Measured reading =`(4 +0.16)mm=4.16mm`
`:.` Absolute reading = Measured reading - e
`=(4.16-0.32)mm=3.84mm`
Therefore, thickness of the glass plate is `3.84 mm`.
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