1. pitch of screw gauge
2.least count of screw gauge
3.the diameter of wire
plz tell with full explanation
Answers
Answer:
Aim
To measure diameter of a given wire using screw gauge.
Materials Required
Screw gauge
Wire
Half-meter scale
Magnifying glass
Theory
What is least count?
Least count is given as:
Leastcount=pitchno.ofdivisionsoncircularscale
If a screw gauge has a pitch of 1 mm and 100 divisions on the circular scale then the least count is given as:
Leastcount=1mm100=0.01mm
How to take the linear scale reading?
Linear scale reading is taken by noting the mark on the linear scale which is close to the edge of the circular scale.
How to take circular scale reading?
Circular scale reading is taken by noting the division on the circular scale that coincides with the main scale reading.
Explanation:
Procedure
Find the value of one linear scale division (L.S.D).
Determine and record the pitch and least count of the screw gauge.
To find the zero error, bring the plane face B and A near. Repeat and record this for three times. Record zero error as nil if there is no error.
Move face B away from face A. using a ratchet head R, move the face A towards face B lengthwise and stop when R turns without moving the screw.
Linear scale reading (L.S.R) is recorded by noting down the no.of visible and uncovered divisions of linear scale.
Let n be the no.of divisions of the circular scale lying on the reference line.
To measure diameter in a perpendicular direction, repeat steps 5 and 6 by rotating the wire to 90°.
For the entire length of wire, repeat steps 4,5,6 and 7 for five different positions and record the observations.
Find the total reading and also zero correction.
Take the mean of different values of diameter.
Using a half-meter scale, measure the length of the wire. Repeat this step three times and record the readings.
Observations
Determination of the least count of the screw gauge L.S.D = 1 mm
Number of full rotations given to screw = 4Distance moved by the screw = 4 mm
Hence, pitch p = 4 mm/4 = 1 mm
No.of divisions on circular scale = 100
Hence, the least count = 1 mm/100 = 0.01 mm = 0.001 cm
Zero error (i)… mm (ii)…. mm (iii)….. mm
Mean zero error (e) =… mmMean zero correction (c) = -e -…. mm
Table for diameter (D)
Answer:
procedure:
Explanation:
Find the value of one linear scale division (L.S.D.).
2 Determine the pitch and the least count of the screw gauge and record it stepwise. 3 Bring the plane face B in contact with plane face A and find the zero error. Do it three times
and record them. If there is no zero error, then record zero error as nil.
4. Move the face B away from face A. Place the wire lengthwise over face A and move the face B towards face A using the ratchet head R. Stop when R turns (slips) without moving the screw.
5. Note the number of divisions of the linear scale visible and uncovered by the edge of the cap.
The reading (N) is called linear scale reading (L.S.R.). 6. Note the number (n) of the division of the circular scale lying over reference line.
7. Repeat steps 5 and 6 after rotating the wire by 90' for measuring diameter in a perpendicular
direction.
53
0.53
(a) * A*Theta*B
53
Pape Me
Dote
D
(b)
53
(a) AB
52
270057
2
54
8. Repeat steps 4, 5, 6 and 7 for five different positions separated equally throughout the length
of the wire. Record the observations in each set in a tabular form.
(b) D
(a) Α Θ Β
9. Find total reading and apply zero correction in each case.
3
10. Take mean of different values of diameter.
09
11. Measure the length of the wire by stretching it along a half-metre scale. Keeping one end of wire at a known mark, note the position of other end. Difference in position of the two ends of the wire gives the length of the wire. Do it three times and record them.