TO READ MEASUREMENT AS SHOWN ABOVE:
First note the whole number of mm divisions on the barrel (major divisions below datum line) ........... 15 x 1,0mm = 15,00mm
Then observe whether there is a half mm visible (minor divisions above datum line)
............3 x 0,50mm = 1,50mm
Finally read the line on the thimble coinciding with the datum line. This gives hundredths of a mm. ........... 16 x 0,01mm = 0,16mm
TOTAL: 15,00mm = 1,50mm 0,16mm = 16,66mm
DIGITAL MICROMETERS
Digital micrometer is very accurate and does involve the computations needed by a standard micrometer. When the spindle and anvil come in contact with the workpiece, the measurement can be read directly from a digital display.
CARE OF MICROMETERS
1. Coat metal parts of all micrometers with a light coat of oil to prevent rust.
2. Store micrometers in separate containers provided by manufacturer.
3. Keep graduations and markings on all micrometers clean and legible.
4. Do not drop any micrometer. Small nicks or scratchescan cause inaccurate measurements.
Question 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 Aluminium. 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 ?
(1) 0.50 mm
(2) 0.75 mm
(3) 0.80 mm
(4) 0.70 mm
Sol.
Least count
= pitch / no. of division on circular scale
= 0.5 mm / 50
= 0.01 mm
Negative zero error = – 5 × LC = -5 × 0.01 = -0.05 mm
Measured value
= main scale reading screw gauge reading – zero error
= 0.5mm 25 × 0.01mm – (-0.05mm)
= 0.75mm 0.05mm
= 0.80mm
Question 2.
Consider a Vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisions and a screw gauge with 100 divisions on its circular scale. In the Vernier callipers, 5 divisions of the Vernier scale coincide with 4 divisions on the main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then:
(A) If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm.
(B) If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm.
(C) If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm.
(D) If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm.
Sol.
Vernier calliper:
M.S.D =1/8 cm = 0.125 cm = 1.25 mm
5 V.S.D. = 4 M.S.D
⇒ V.S.D. = 4/5 M.S.D. = 4/5 × 1.25mm = 1 mm
Least count = M.S.D. – V.S.D = 1.25 – 1 = 0.25 mm
Screw gauge:
No. of circular scale divisions = 100
1 rotation (pitch) = 2 linear scale divisions
» If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is –
Pitch = 2 × Least Count of vernier callipers = 2 × 0.25mm = 0.5 mm
Least count of the screw gauge
= Pitch/no. of circular scale divisions = 0.5/100 =0.005mm
Option (B) is correct.
» If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is –
Linear scale division of screw gauge
= 2 × Least count of vernier callipers
= 2 × 0.25 mm
= 0.5 mm
Now,
Pitch = 2 linear scale divisions = 2 × 0.5 = 1mm
Least count of the screw gauge
= Pitch/ No. of circular divisions
= 1/100 mm
= 0.01 mm
Option (C) is correct.
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