Problems on Screw Gauge
Example 1
A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminum. Before starting the measurement , it is found that when the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line ?
0.50 mm
0.75 mm
0.80 mm
0.70 mm

We know that
?ℎ???????,?=???+???×??(????? ???? ?????????? ?? ???)
???=0.5??
???=25?ℎ ???????
??=????ℎ/(?? ?? ????????)=0.5/50=0.01??
???? ??????????=5×0.01
?=0.5??+25∗0.01??+0.05??=0.80??
Example 2
[JEE 2004]
The pitch of a screw gauge is 1 mm and there are 100 divisions on the circular scale. While measuring the diameter of a wire, the linear scale reads 1 mm and 47th division on the circular scale coincides with the reference line. The length of the wire is 5.6 cm. Find the curved surface area (in cm) of the wire in appropriate number of significant figures.

We know that
????????,?=???+???×??(????? ???? ?????????? ?? ???)
?=1??+47∗1/100=1.47??=0.147??
?=???=3.14∗0.147∗5.6=2.5848??
?=2.6??



Example 3

[JEE 2011]
The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale. The reading on the main scale is 2.5 mm and that on the circular scale is 20 divisions. If the measured mass of the ball has relative error of 2%, the percentage error in density is
0.9%
2.4%
3.1%
4.2%

We know that
????????,?=???+???×??(????? ???? ?????????? ?? ???)
?=2.5??+20∗0.5/50 ??=2.7??
??=??=0.01??
??/?×100=1/2.7

?=?/?=?/(1/6 ??^3 )=6?/(??^3 )
ln⁡〖?=ln⁡6 〗+ln⁡?−ln⁡?−3 ln⁡?
??/?=??/?−3 ??/?
??/?×100=??/?×100+3 ??/?×100


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