Measurement tools and techniques
Explore the use and principles of common measuring instruments like vernier calipers and screw gauge, including least count, zero error, and practical measurement methods.
Why this shows up in the exam
NEET tests your ability to interpret readings and correct errors in measurements using standard laboratory instruments.
How NEET tests this
Learn the idea
A screw gauge or vernier calipers measures small lengths by reading the main scale and the finer circular/vernier scale. The true measurement is obtained after correcting for the instrument's least‑count and any zero error.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- Zero error = (error divisions) × least‑count; add if negative, subtract if positive
- Reading = main‑scale reading + circular (or vernier) scale reading – zero error
- For a screw gauge, diameter = MSR + CSD – zero error; radius = half of the diameter
How to approach it
- 1Determine the least‑count from pitch and divisions
- 2Find the zero error from the position of the zero on the circular scale and compute its magnitude
- 3Add the main‑scale reading and the circular (or vernier) reading, then subtract the zero error to obtain the true value
Worked example — watch it click
The pitch of a screw gauge is 1 mm and there are 100 divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lies 8 divisions below the reference line. When a wire is placed between the jaws, the first linear scale divisions is clearly visible while the 72nd division on a circular scale coincides with reference line. The radius of the wire is
- A)1.64 mm
- ✅0.82 mm
- C)1.80 mm
- D)0.90 mm
The concept behind this problem
The example forces you to calculate least‑count, identify a positive zero error, and apply the reading formula, thereby testing the complete procedure for a screw‑gauge measurement.
Step by step
- 1Pitch = 1 mm, circular scale has 100 divisions
- 2least-count = 1 mm/100 = 0.01 mm per division. When the jaws are open the zero of the circular scale is 8 divisions below the reference line, so the instrument has a positive zero error of +8 × 0.01 mm = +0.08 mm (the reading will be larger than the true value). With the wire in the jaws the main-scale reading is the first division
- 31 mm, and the 72nd circular division coincides with the reference line
- 472 × 0.01 mm = 0.72 mm. Raw reading (diameter shown) = 1 mm + 0.72 mm = 1.72 mm. Corrected diameter = raw reading – zero error = 1.72 mm − 0.08 mm = 1.64 mm. Radius = (corrected diameter)/2 = 1.64 mm / 2 = 0.82 mm. So the correct answer is 0.82 mm.
Watch out
A positive zero error is always subtracted, not added, to get the correct dimension.
Common slip-ups that cost marks
- •Missing the sign of zero error (positive error must be subtracted)
- •Confusing diameter with radius or mixing mm and cm units
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Practise it
These are real questions from past NEET papers that test this exact idea.
In a vernier callipers, 20 VSD coincide with 16 MSD (each division of length 1 mm). The least count of the vernier callipers is:
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