A screw gauge with a pitch of 0.5 mm and has 50 divisions on its circular scale. Its linear and circular scale readings are respectively, 2mm and 2 s when the diameter of a sphere is measured by the gauge. The result of this measurement is
Answers
As we learnt in
To measure the thickness of the given sheet using screw gauge -
1. Note the number of divisions on the circular scale.
2. Give five complete rotations to the screw.
3. Note the linear distance moved by the screw.
4. Find the pitch and L.C. of screw gauge.
5. Find the zero error and zero correction by moving the screw only in one direction in such a way that studs A and B just touch each other.
6. Now grip the given sheet in the gap A and B of the screw gauge.
7. Turn the screw head till the ratchet arrangement gives a click.
8. Note the readings of linear scale and circular scale and find the observed thickness using the relation, observation thickness = L.S.R. + C.S.R.
9. Add the zero correction to the observed thickness to find the corrected diameter.
10. Repeat steps 6 to 9 to find the thickness from four more different places.
- wherein
C.S.R= n \times L.C
Total reading
= L.S.R + C.S.R
= N\times n +N.C.
L.C = least count
C.S.R = Circular scale reading
N= Nth division of linear scale
Least \: count=\frac{pitch}{no. \: of \: division \: on \: circular \: scale}
=\frac{0.5mm}{50}
L.C.=0.001mm
-ve \ zero \: error=-5\times LC= -0.005mm
Measured value=main scale reading + screw gauge reading - zero error
=0.5mm+(25\times 0.001-(-0.05))mm=0.80mm
Option 1)
0.75 mm
This option is incorrect
Option 2)
0.80 mm
This option is correct
Option 3)
0.70 mm
This option is incorrect
Option 4)
0.50 mm
This option is incorrect