Vernier caliper and screw gauge — reading techniques and error calculation

Question

How do you take a reading using a Vernier caliper and a screw gauge? What is the least count of each instrument? How do you calculate zero error and apply the correction? Solve a numerical example for each.

(CBSE 11 + JEE Main + NEET — practical + numerical)

Solution — Step by Step

Least count = 1 MSD - 1 VSD

For a standard Vernier caliper: $\text{LC} = 1 \text{ mm} - 0.9 \text{ mm} = 0.1 \text{ mm} = 0.01 \text{ cm}$

Reading = Main Scale Reading (MSR) + Vernier Scale Reading (VSR) $\times$ LC

\text{Reading} = \text{MSR} + (\text{VSD coinciding}) \times 0.01 \text{ cm}

Example: MSR = 3.2 cm, 6th VSD coincides with a main scale division.

\text{Reading} = 3.2 + 6 \times 0.01 = 3.26 \text{ cm}

Least count = Pitch / Number of divisions on circular scale

For a standard screw gauge: $\text{LC} = \frac{1 \text{ mm}}{100} = 0.01 \text{ mm} = 0.001 \text{ cm}$

Reading = Linear Scale Reading (LSR) + Circular Scale Reading (CSR) $\times$ LC

\text{Reading} = \text{LSR} + (\text{circular scale division}) \times 0.001 \text{ cm}

Example: LSR = 4.5 mm, CSR = 32

\text{Reading} = 4.5 + 32 \times 0.01 = 4.82 \text{ mm}

Type	What You See	Correction
Positive zero error	Zero of Vernier/circular scale is AHEAD of zero of main scale	Subtract zero error from reading
Negative zero error	Zero of Vernier/circular scale is BEHIND zero of main scale	Add the correction (total divisions - error reading) $\times$ LC
No zero error	Both zeros coincide perfectly	No correction needed

Corrected reading = Observed reading - Zero error

(where positive zero error is +ve and negative zero error is -ve, so subtracting a negative error effectively adds it)

A screw gauge has a positive zero error of 3 divisions. LC = 0.01 mm.

Zero error = $+3 \times 0.01 = +0.03$ mm

Observed reading: LSR = 2.5 mm, CSR = 47

Observed = $2.5 + 47 \times 0.01 = 2.97$ mm

Corrected = $2.97 - 0.03 = \mathbf{2.94 \text{ mm}}$

graph TD
    A["Measurement Reading"] --> B["Read Main/Linear Scale"]
    B --> C["Read Vernier/Circular Scale"]
    C --> D["Observed = MSR + VSR x LC"]
    D --> E{"Zero error?"}
    E -->|Positive| F["Corrected = Observed - ZE"]
    E -->|Negative| G["Corrected = Observed + |ZE|"]
    E -->|None| H["Corrected = Observed"]
    style A fill:#fbbf24,stroke:#000,stroke-width:2px
    style D fill:#86efac,stroke:#000

Why This Works

Both instruments use a secondary scale to subdivide the smallest division of the primary scale. The Vernier caliper uses a sliding scale where slightly smaller divisions create a magnification effect. The screw gauge uses a rotating thimble — one full rotation advances the spindle by one pitch, and the circular scale divides this pitch into 50 or 100 parts.

Zero error occurs due to manufacturing imperfections — the jaws or spindle do not align perfectly when fully closed. Knowing how to detect and correct zero error is essential for accurate measurements.

Common Mistake

The most common error: adding positive zero error instead of subtracting it. If the instrument reads too high (positive zero error), the true value is LESS than the observed reading — so you SUBTRACT. Think of it this way: if your weighing scale shows 2 kg when nothing is on it (positive zero error of 2 kg), you must subtract 2 from every reading. The same logic applies to Vernier and screw gauge.

For JEE/NEET: remember that Vernier caliper LC = 0.01 cm and screw gauge LC = 0.01 mm (or 0.001 cm). The screw gauge is 10x more precise. Questions asking “which instrument is more accurate?” always expect screw gauge. Also, “least count” = smallest measurable value = instrument precision — not accuracy.

Question

Solution — Step by Step

Why This Works

Common Mistake

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