Class 11 Physics practical Vernier Calliper

Aim

(i) To measure the diameter of a given spherical body (e.g. a glass marble) using Vernier Callipers

(ii) To measure the internal diameter and depth of a given beaker using Vernier Callipers, and hence to determine its volume.

Apparatus and Material Required

• Vernier Callipers
• A spherical body (e.g. a glass marble)
• A beaker
• A magnifying lens (optional, for reading the vernier scale clearly)

Principle

A Vernier Callipers measures lengths more precisely than an ordinary scale. It has a fixed main scale (graduated in cm/mm) and a sliding vernier scale. It carries two pairs of jaws: the lower, external jaws measure outer dimensions (e.g. the diameter of a sphere, or the outside of a beaker); the upper, internal jaws measure an internal width, such as the mouth of a beaker. A thin depth probe, attached to the sliding part, extends out the back and measures the depth of a hollow object.

Least count (L.C.):

L.C. = 1 M.S.D. − 1 V.S.D. = (value of 1 M.S.D.) / N
where M.S.D. = main scale division, V.S.D. = vernier scale division, N = total divisions on the vernier scale.

Zero error: With the jaws fully closed, the vernier zero should coincide exactly with the main scale zero. If it doesn’t, the instrument has a zero error, which must be applied (with the correct sign) to every reading.

• Positive zero error — vernier zero lies to the right of the main scale zero: error = +(n × L.C.), where n is the coinciding vernier division.
• Negative zero error — vernier zero lies to the left of the main scale zero: error = −(N − n) × L.C.

Corrected reading = Observed reading − Zero error

Formulae Used

Volume of a spherical body of diameter D: V = (4/3)πr³ = πd³/6
Volume of the beaker/calorimeter (internal diameter D, internal depth h): V = πD²h/4

Procedure

• Least count and zero error
  1. Note the value of 1 main scale division and the total number of vernier divisions, N; calculate the least count.
  2. Close the jaws gently and check whether the vernier zero coincides with the main scale zero. Note the type and magnitude of zero error, if any; repeat three times and take the mean.
• Diameter of the spherical body
  1. Grip the body gently but firmly between the external (lower) jaws so it doesn’t slip, keeping the jaws perpendicular to the diameter.
  2. Note the main scale reading (MSR) just before the vernier zero, and the vernier division that exactly coincides with a main scale division.
  3. Repeat along 3–4 different directions of the body to check uniformity, and tabulate every reading.
• Internal diameter of the beaker
  1. Insert the internal (upper) jaws into the mouth of the vessel and open them gently until they just touch the inner walls, perpendicular to the diameter.
  2. Note the MSR and coinciding vernier division; repeat at two mutually perpendicular positions, 3–4 readings in all.
• Internal depth of the beaker
  1. Rest the main scale’s edge flat on the rim and slide the depth probe down until it just touches the bottom, exactly perpendicular to the base.
  2. Note the MSR and coinciding vernier division; repeat at 3–4 different points on the bottom.

Apply the zero correction to every observed reading before taking means and substituting into the volume formulae.

Observations

Table 1 — Least Count

Quantity	Value
Value of 1 main scale division (1 M.S.D.)	1 mm
Number of divisions on vernier scale, N	10
10 vernier scale divisions	9 M.S.D.
1 vernier scale division	0.9 mm
Least count = 1 M.S.D. − 1 V.S.D.	0.1 mm
Zero error observed (mean of 3 trials):	Nil (0.0 mm)

Table 2 — Diameter of the spherical body

Trial	MSR, a (mm)	Coinciding V.S.D., n	n × L.C. (mm)	Diameter = a + n×L.C. (mm)
1.	12	4	0.4	12.4
2.	12	5	0.5	12.5
3.	12	3	0.3	12.3
4.	12	4	0.4	12.4

Mean diameter, d = 12.4 mm

Table 3 — Internal diameter of the beaker

Trial	MSR, a (mm)	Coinciding V.S.D., n	n × L.C. (mm)	Diameter = a + n×L.C. (mm)
1.	41	5	0.5	41.5
2.	42	4	0.4	42.4
3.	42	5	0.5	42.5
4.	42	6	0.6	42.6

Mean internal diameter, D = 42.2 mm

Table 4 — Internal depth of the beaker

Trial	MSR, a (mm)	Coinciding V.S.D., n	n × L.C. (mm)	Diameter = a + n×L.C. (mm)
1.	49	6	0.6	49.6
2.	50	7	0.7	50.7
3.	50	4	0.4	50.4
4.	50	3	0.3	50.3

Mean internal depth, h = 50.2 mm

Calculations

Volume of the spherical body:
V = πd³/6 = 3.14 × (12.4)³ / 6 = 3.14 × 317.8 mm³ = 998 mm³ ≈ 1.00 cm³

Volume of the beaker:
V = πD²h/4 = 3.14 × (42.5)² × 52.2 / 4 = 74014.7 mm³≈ 74.01 cm³

Results

Volume of the spherical body: 1.00 cm³

Volume of the beaker: 74.01 cm³

Precautions

1. Move the jaws gently by turning the screw; never force them. The jaws should grip the body firmly enough to prevent slipping, but not so tightly as to compress it.
2. Take readings with the eye perpendicular to the scale, to avoid parallax error.
3. Note the zero error carefully before starting, and apply correction with the proper sign to every reading.
4. Measure the diameter along several directions to check uniformity.
5. While measuring internal diameter, the jaws should just touch the walls and be exactly perpendicular to the axis of the vessel.
6. While measuring depth, the strip must be exactly perpendicular to the base, and the main scale’s edge must rest flat on the rim.

Sources of Error

• Excessive jaw pressure on a soft or fragile body compresses it, giving a smaller reading than the true dimension.
• Wear and backlash in the sliding vernier scale from prolonged use.
• Parallax while reading the coinciding vernier division.
• Non-uniform shape of the given body or vessel. Manufacturing non-uniformity in the scale’s graduations.

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Reference: NCERT Lab Manual