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Physics - Measurements and Experimentation (Vernier Callipers, Screw Gauge)

Grade 9ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Least Count (LCLC) is the smallest measurement that can be taken accurately with a measuring instrument.

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A Vernier Calliper consists of a Main Scale and a Vernier Scale. The LCLC is the difference between the value of 11 Main Scale Division (MSDMSD) and 11 Vernier Scale Division (VSDVSD).

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A Screw Gauge works on the principle of a screw. The distance moved by the spindle in one complete rotation is called the PitchPitch.

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Zero Error occurs when the zero mark of the Vernier scale or Circular scale does not coincide with the zero mark of the Main scale/Base line when the jaws or studs are in contact.

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Positive Zero Error occurs when the Vernier/Circular zero is ahead of (to the right of/above) the Main zero. It must be subtracted from the observed reading.

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Negative Zero Error occurs when the Vernier/Circular zero is behind (to the left of/below) the Main zero. It must be added to the observed reading (or subtracted as a negative value).

📐Formulae

LCVernier=Value of 1 Main Scale DivisionTotal number of divisions on Vernier ScaleLC_{\text{Vernier}} = \frac{\text{Value of 1 Main Scale Division}}{\text{Total number of divisions on Vernier Scale}}

Total Reading (Vernier)=MSR+(VSC×LC)\text{Total Reading (Vernier)} = MSR + (VSC \times LC) where MSRMSR is Main Scale Reading and VSCVSC is Vernier Scale Coincidence.

Pitch=Distance moved on Main ScaleNumber of full rotations given to the thimblePitch = \frac{\text{Distance moved on Main Scale}}{\text{Number of full rotations given to the thimble}}

LCScrew Gauge=PitchTotal number of divisions on the Circular ScaleLC_{\text{Screw Gauge}} = \frac{Pitch}{\text{Total number of divisions on the Circular Scale}}

Total Reading (Screw Gauge)=MSR+(CSR×LC)\text{Total Reading (Screw Gauge)} = MSR + (CSR \times LC) where CSRCSR is Circular Scale Reading (coinciding division).

Correct Reading=Observed Reading−(±Zero Error)\text{Correct Reading} = \text{Observed Reading} - (\pm \text{Zero Error})

💡Examples

Problem 1:

In a Vernier Calliper, the main scale is graduated in mmmm. 1010 vernier divisions coincide with 99 main scale divisions. If the main scale reading is 1.3 cm1.3\text{ cm} and the 7th7^{th} vernier division coincides with a main scale division, find the total reading.

Solution:

  1. 1 MSD=1 mm=0.1 cm1\text{ MSD} = 1\text{ mm} = 0.1\text{ cm}.
  2. LC=1 MSD10=0.1 cm10=0.01 cmLC = \frac{1\text{ MSD}}{10} = \frac{0.1\text{ cm}}{10} = 0.01\text{ cm}.
  3. MSR=1.3 cmMSR = 1.3\text{ cm} and VSC=7VSC = 7.
  4. Total Reading=1.3+(7×0.01)=1.37 cm\text{Total Reading} = 1.3 + (7 \times 0.01) = 1.37\text{ cm}.

Explanation:

The least count is calculated first, then the fractional part (Vernier Reading) is added to the Main Scale Reading.

Problem 2:

A screw gauge has a pitch of 1.0 mm1.0\text{ mm} and 100100 divisions on its circular scale. While measuring the diameter of a wire, the main scale reads 2 mm2\text{ mm} and the 45th45^{th} mark on the circular scale coincides with the base line. Find the diameter.

Solution:

  1. LC=PitchNo. of divisions=1.0 mm100=0.01 mmLC = \frac{Pitch}{\text{No. of divisions}} = \frac{1.0\text{ mm}}{100} = 0.01\text{ mm}.
  2. MSR=2 mmMSR = 2\text{ mm} and CSR=45CSR = 45.
  3. Diameter=MSR+(CSR×LC)=2+(45×0.01)=2.45 mm\text{Diameter} = MSR + (CSR \times LC) = 2 + (45 \times 0.01) = 2.45\text{ mm}.

Explanation:

The pitch divided by circular scale divisions gives the precision (LCLC), which is then multiplied by the coinciding division to find the circular scale contribution.