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Construction:
 Vernier callipers is device is used to measure the internal and external diameter of a tube, the external diameter of a sphere, depth of a vessel, the diameter of thick wires and cylinders.
 Typical vernier callipers consist of a steel strip which is generally marked in centimetre and millimetre along its lower edge this is fixed scale called the main scale. The end of the main scale is provided with a fixed jaw forming external jaw on the lower side and internal jaw on the upper side.
 A sliding frame with a graduation on lower side slides over the main scale. This sliding scale is called vernier scale. The inside end of a vernier scale is provided with a fixed jaw forming external jaw on the lower side and internal jaw on the upper side.
 Generally, the vernier scale is provided with N divisions matching with (N – 1) divisions of the main scale.
Least Count of Vernier Scale:
 The minimum length that can be measured using the vernier callipers is called its least count.
Errors:

Zero Errors of Vernier Calipers:
 When the two jaws of vernier callipers are made to touch each other, then the zero on the main scale should match with a zero on the vernier scale.
 However due to wear and tear or manufacturing defect the two zeros usually do not coincide with each other, then the vernier is said to have zero error.
 There are two types of zero errors.

Positive zero error:
 If on bringing both the jaws together, the zero mark of the vernier scale is on the right side of a zero mark of the main scale, then the zero error is said to be positive.
 To find positive zero error, note the division on the vernier scale (V.S.R.) coinciding with the division on the main scale. Then the positive error is equal to the product of the number of division on vernier scale matching with main scale division and the least count of the vernier.
 To get correct reading this error is to be subtracted from overall reading.

Negative zero error:
 If on bringing both the jaws together, the zero mark of the vernier scale is on the left side of a zero mark of the main scale, then the zero error is said to be negative.
 To find negative zero error, note the division on the vernier scale (V.S.R.) coinciding with the division on the main scale. Subtract this number from a total number of divisions on the vernier scale to obtain V.S.R. Then the negative error is equal to the product of V.S.R. and the least count of the vernier.
 To get correct reading this error is to be added to overall reading.
Use of Vernier Calipers:
 The object whose dimensions to be measured is held in jaws of the callipers with gentle pressure.
 Note down the main scale reading just before zero of vernier callipers. This is called main scale reading (M.S.R.)
 Note down the number of vernier scale division (n) which coincides with any division of the main scale. Then vernier scale reading, (V.S.R.) = n X Least Count.
 Add the M.S.R. and the V.S.R. to get reading.
 Subtract the zero error with a proper sign from above reading to get the correct reading.
Uses of Different Parts of Vernier Calipers:
 The outside jaw is used to measure outside dimensions like the length of a rod, the external diameter of sphere or cylinder.
 The inside jaw is used to measure inside dimensions like internal diameter of ring or hollow cylinder or pipe
 The strip is used to measure the depth of a beaker or bottle.
 The main scale is used to measure the length of an object correct up to 1 mm.
 Vernier scale is used to measure length of an object correct up to 0.1 mm
Dial Vernier Calipers (Direct Reading)
Digital Vernier Calipers (Direct Reading)
Example 1:
 The main scale of a vernier scale has least count of 0.5 mm. If 20 divisions of this scale are divided into 25 equal parts of vernier scale, what is the least count of the vernier calliper?
 Solution:
1 M.S.D. = 0.5 mm
25 V.S.D. = 20 M.S.D.
Thus, 1 V.S.D. = 20/25 M.S.D. = 0.8 M.S.D. = 0.8 x 0.5 mm = 0.4 mm
Now, Least Count of Vernier Caliper = 1 M.S.D. – 1.V.S.D. = 0.5 mm – 0.4 mm = 0.1 mm
Hence least count of vernier calliper is 0.1 mm
Example – 2:
 On vernier callipers, one centimetre of the main scale is divided into 10 equal parts. If 10 divisions of vernier scale coincide with 9 divisions of the main scale, then what will be the least count of the vernier calliper.
 Solution:
1 M.S.D. = cm = 0.1 cm
10 V.S.D. = 9 M.S.D.
Thus, 1 V.S.D = 9/10 M.S.D. = 0.9 M.S.D. = 0.9 x 0.1 cm = 0.09 cm
Now, Least Count of Vernier Caliper = 1 M.S.D. – 1.V.S.D. = 0.1 cm – 0.09 cm = 0.01 cm
Hence least count of vernier calliper is 0.01 cm
Example – 3:
 On vernier callipers, one centimetre of the main scale is divided into 20 equal parts. If the number of divisions on vernier scale is 25. Find the least count of the vernier calliper. In a measurement of the length of an object, the main scale reading lies between 2.35 cm and 2.40 cm. Vernier scale reading is 6. Find the length of the object.
Solution:
On main scale 1 cm is divided into 20 equal parts. Hence 1 M.S.D. = 1/20 cm = 0.05 cm.
Now least count = (1/No. of divisions of vernier scale) x M.S.D. = (1/25) x 0.05 cm = 0.002 cm.
Reading = Main scale reading + Vernier scale reading x Least Count Reading = 2.35 + 6 x 0.002 = 2.35 + 0.012 = 2.362 cm
Hence the length of the object is 2.362 cm
Example – 4:
 On vernier callipers, one centimetre of the main scale is divided into 20 equal parts. If the number of divisions on vernier scale is 25. Find the least count of the vernier calliper. In a measurement of the length of an object, the main scale reading lies between 2.75 cm and 2.80 cm. Vernier scale reading is 9. The zero error is + 0.024 cm. Find the length of the object.
Solution:
On main scale 1 cm is divided into 20 equal parts. Hence 1 M.S.D. = 1/20 cm = 0.05 cm.
Now least count = (1/No. of divisions of vernier scale) x M.S.D. = (1/25) x 0.05 cm = 0.002 cm.
Reading = Main scale reading + Vernier scale reading x Least Count – Zero error Reading
= 2.75 + 9 0.002 – (+ 0.024) = 2.75 + 0.018 – 0.024 = 2.744 cm
Hence the length of the object is 2.744 cm
Example – 5:
 On vernier callipers, one centimetre of the main scale is divided into 10 equal parts. If the number of divisions on vernier scale are 10. Find the least count of the vernier calliper. In a measurement of the length of an object, the main scale reading lies between 5.6 cm and 5.7 cm. Vernier scale reading is 4. The zero error is – 0.02 cm. Find the length of the object.
Solution:
On main scale 1 cm is divided into 10 equal parts. Hence 1 M.S.D. = 1/10 cm = 0.1 cm.
Now least count = (1/No. of divisions of vernier scale) x M.S.D. = (1/10) x 0.1 cm = 0.01 cm.
Reading = Main scale reading + Vernier scale reading Least Count – Zero error Reading
= 5.6 + 4 0.01 – ( 0.02) = 5.6 + 0.04 +0.02 = 5.66 cm
Hence the length of the object is 5.66 cm
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Science > Physics > Units & Measurements > You are Here 
Physics  Chemistry  Biology  Mathematics 
bravo!!!, really helping thanks
Plz give a zooming view of main and ver scale.It would be easy to understand.Thanx
Added the zooming view as per your suggestion. Thank you for your suggestion.
I can’t understand why should we take the lower main scale reading when given within a range.
We have started adding solved question bank for HSC Maharashtra Board the link is https://hemantmore.org.in/physics/maharashtrastateboardquestionbankphysics/