Physics · Units and Measurements · Experimental Skills

Vernier Caliper and Screw Gauge Simulator

Move the slider to change the object's length, add a zero error, and read the instrument exactly as JEE asks.

Last updated: 26 October 2026

Direct answer

Reading = main scale reading + (coinciding division × least count) − zero error.

A vernier caliper with 10 divisions equal to 9 mm has a least count of 0.1 mm. A screw gauge with 0.5 mm pitch and 50 circular divisions has a least count of 0.01 mm. Positive zero error is subtracted; negative zero error is added.

Interactive

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📏 Vernier and Screw Gauge

Experimental Skills
Preparing simulation…The explanation below is already readable — the interactive canvas starts in a moment.
What's happening? Find the vernier (or circular-scale) division that lines up with a main-scale mark. Multiply it by the least count, add the main scale reading, then correct for zero error.
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Worked example

Worked example

  • Vernier: MSR = 23 mm, 4th division coincides, LC = 0.1 mm → 23 + 0.4 = 23.4 mm.
  • Zero error +0.2 mm → corrected reading 23.2 mm.
  • Screw gauge: MSR = 5.5 mm, circular reading 27, LC = 0.01 mm → 5.5 + 0.27 = 5.77 mm.
Avoid

Common mistakes

  • Adding a positive zero error instead of subtracting it
  • Using 1 mm instead of 0.5 mm as the screw gauge main-scale step
  • Forgetting to count a negative zero error as (n − total divisions)
FAQs

Frequently asked questions

What is the least count of a vernier caliper?

Least count = 1 main scale division − 1 vernier scale division. For 10 vernier divisions equal to 9 mm, it is 0.1 mm.

How do you correct zero error?

Corrected reading = observed reading − zero error. A positive zero error is subtracted and a negative one is added.

What is the least count of a screw gauge?

Pitch ÷ number of circular divisions. For 0.5 mm pitch and 50 divisions, it is 0.01 mm.

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