Uploaded by Jay C Cooper

LAB REPORT-1A

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Experiment 1
The Screw Gauge
Aim:To find (a) the diameter of a wire, and (b) volume of a spherical ball using screw
gauge.
Apparatus:Screw gauge, the wire, and the spherical ball.
The screw gauge consists of a very fine screw with a flat tip working in a groove
cut inside a hollow cylinder. The hollow cylinder is fitted to a U shaped frame.
Parallel to the axis of the cylinder, a scale graduated in mm is engraved. This is
called the pitch scale. A sleeve is attached to the head of the screw. The head of
the screw has a ratchet which avoids undue frightening of the screw. The edge
of the sleeve is divided into equal parts (usually 50 or 100 equal divisions). This
is known as the head scale. When the screw is worked, the sleeve moves over
the pitch scale. A stud with a plane end surface is fixed on the “U” frame exactly
opposite to the tip of the screw. When the tip of the screw is in contact with the
stud, usually, the zero of the head scale coincides with the zero of the pitch scale.
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Theory:Dimension measured = P.S.R + fraction
= P.S.R + (H.S.R x L.C)
Where, P.S.R, the pitch scale reading, H.S.R, the head scale reading and L.C, the
least count.
Procedure:(a) To find the pitch of the screw:The pitch is the distance moved by the screw for one complete rotation of the
head of the screw. To find this, the distance advanced by the head scale over the
pitch scale for a definite number of complete rotation of the screw is determined.
From this, the pitch of the screw is calculated.
Pitch =Distance moved/ Number of rotations
(b) To find the least count of the screw gauge
Least count = Pitch/ Number of divisions on the head scale
( c) To find zero correction
The screw is rotated till the tip just touches the stud. If the zero of
the head scale does not coincide with the zero of the pitch scale, there is a zero
error. So a correction has to be applied.
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Observations:-
Distance moved for 5 rotations
Pitch of the screw
2.5
=…………………mm
0.5
=………………….mm
50
No. of divisions on the head scale = ……………………
Least count
1.0 x 10-5
0.01
= ………………….
mm =……………………..
m
Zero correction
-25
= ………………………..
divisions
Diameter of the wire
Trial
P.S.R (mm)
Observed
H.S.R
Corrected
H.S.R
Fraction=
Corrected
H.S.R x
L.C (mm)
Diameter
of the
wire =
P.S.R
+Fraction
(mm)
0.5
14.0
-11.0
-0.110
0.390
0.5
14.5
-10.5
-0.105
0.395
0.5
15.0
-10.0
-0.100
0.400
0.5
15.0
-10.0
-0.100
0.400
0.5
14.5
-10.5
-0.105
0.395
0.5
14.5
-10.5
-0.105
0.395
1.
2.
3.
4.
5.
6.
0.396
Mean diameter of the wire =…………………………mm
0.198
1.98 x 10-4
Mean radius of the wire =………………..mm
= …………………………m
0.3850
38.50
Length of the wire; l= ………………cm=
…………………………….m
3
4.74 x 10-8
Volume of the wire =………………………m
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