Vernier callipers measure linear dimensions to 0.01 cm. Least count = 1 main-scale division − 1 vernier-scale division. Zero error (positive/negative) is determined by closing the jaws and noted; readings are corrected by subtracting the zero error. Used to measure internal/external diameter and depth of a vessel.
-- NCERT Physics Lab Manual Class 11, Part 2, p. 24Experiment 01 Vernier Calipers
8 MCQs2 revision cards9-step worked example
Source: NCERT Experimental SkillsOfficial key: NTA-verifiedLast updated: 25 Sep 2026
Experiment 01 Vernier Calipers, explained for NEET
The reading you write down is almost never the reading the caliper gave you. Two corrections stand between them, and both are commonly dropped.
The first is zero error. Close the jaws. If the vernier zero does not sit exactly on the main-scale zero, every measurement you take carries that same offset. Vernier zero to the right of the main zero is a positive zero error and is subtracted; vernier zero to the left is a negative zero error, found as −(N − n) × LC where n is the coinciding division, and is added back. The sign is the whole trap — flipping it moves the answer by twice the error, which is usually enough to land on a distractor.
The second is precision. The least count fixes how many digits you are entitled to write: LC = 1 MSD − 1 VSD = 1 MSD / N. For the standard instrument with 1 MSD = 1 mm and N = 10, LC = 0.1 mm = 0.01 cm, so the reading ends at the second decimal in centimetres. A calculator mean of 2.5733 cm is still a 0.01 cm measurement; reporting 2.5733 cm claims a resolution the instrument does not have.
The procedure itself is in the NCERT Physics Lab Manual Class 11, Part 2, page 24: the lower external jaws take the external diameter, the upper internal jaws take the internal diameter, and the thin strip sliding out of the far end of the main scale takes the depth. Total reading = main-scale reading + (coinciding vernier division × LC), then apply the zero-error correction.
Watch-out for NEET: a stem that says "when the jaws are closed the reading is …" is asking for the sign, not the arithmetic. Decide the direction of the offset before you touch the numbers.
Can you answer these Experiment 01 Vernier Calipers MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
The least count of a measuring instrument is defined as:
Show answer and why every option is right or wrong
Answer: C. C is correct — least count is the smallest quantity the instrument can resolve, as defined for the vernier callipers in the NCERT Physics Lab Manual Class 11, Part 2, page 24.
Why A is wrong: A is wrong because that is the instrument's range, not its resolution; a 15 cm caliper and a 30 cm caliper can share an LC of 0.01 cm.
Why B is wrong: B is wrong because the gap between two successive main-scale marks is 1 MSD itself; the least count is 1 MSD − 1 VSD, which is smaller.
Why D is wrong: D is wrong because no averaging is involved; LC comes from the difference 1 MSD − 1 VSD, equivalently 1 MSD / N.
MCQ 2Easy RecallPractice
With the jaws of a vernier calliper fully closed, a positive zero error is present when the vernier zero lies:
Show answer and why every option is right or wrong
Answer: A. A is correct — a vernier zero displaced to the right of the main-scale zero on closed jaws is a positive zero error, which is subtracted from every observed reading (NCERT Physics Lab Manual Class 11, Part 2, page 24).
Why B is wrong: B is wrong because a vernier zero to the left is a negative zero error, which is added back rather than subtracted (trap: zero-error sign confusion).
Why C is wrong: C is wrong because coincidence of the two zeros is exactly the case of no zero error.
Why D is wrong: D is wrong because the closed-jaw position places the vernier zero near the main-scale zero; the far end of the scale is not involved.
MCQ 3Easy RecallPractice
The depth of a beaker is measured with a vernier calliper using:
Show answer and why every option is right or wrong
Answer: D. D is correct — the thin depth strip attached to the sliding vernier frame emerges from the end of the main scale and rests on the vessel's inner base, as described in the NCERT Physics Lab Manual Class 11, Part 2, page 24.
Why A is wrong: A is wrong because the lower (external) jaws grip an object from outside and cannot reach the inner base of a vessel.
Why B is wrong: B is wrong because the upper (internal) jaws open outwards against the inner walls and give the internal diameter, not the depth.
Why C is wrong: C is wrong because closed jaws give the zero-error check, not a depth.
MCQ 4Direct ApplicationPractice
On a vernier calliper the smallest main-scale division is 1 mm, and 20 vernier divisions coincide in length with 19 main-scale divisions. The least count is:
Show answer and why every option is right or wrong
Answer: B. B is correct — 20 VSD = 19 MSD gives 1 VSD = 0.95 mm, so LC = 1 MSD − 1 VSD = 0.05 mm, the same as 1 MSD / N = 1/20 mm (NCERT Physics Lab Manual Class 11, Part 2, page 24).
Why A is wrong: A is wrong because 0.5 mm is half a main-scale division and would follow only from N = 2; the vernier here has 20 divisions.
Why C is wrong: C is wrong because 0.02 mm comes from N = 50; using the wrong division count is the usual slip here.
Why D is wrong: D is wrong because 0.1 mm is the LC of the standard 10-division vernier, not of this 20-division one.
MCQ 5Direct ApplicationPractice
A vernier calliper of least count 0.01 cm shows, with the jaws closed, the 3rd vernier division coinciding with a main-scale mark while the vernier zero lies to the right of the main-scale zero. A rod then gives a main-scale reading of 2.4 cm with the 6th vernier division coinciding. The corrected diameter of the rod is:
Show answer and why every option is right or wrong
Answer: C. C is correct — observed reading = 2.4 + 6 × 0.01 = 2.46 cm, and the positive zero error of +0.03 cm is subtracted, giving 2.43 cm (NCERT Physics Lab Manual Class 11, Part 2, page 26).
Why A is wrong: A is wrong because it adds the zero error instead of subtracting it; the vernier zero sitting to the right means the instrument over-reads (trap: zero-error sign confusion).
Why B is wrong: B is wrong because 2.46 cm is the raw observed reading with no zero-error correction applied at all.
Why D is wrong: D is wrong because it drops the vernier contribution 6 × 0.01 cm and keeps only the main-scale reading.
MCQ 6Direct ApplicationPractice
Using a vernier calliper of least count 0.01 cm, a student averages three readings of a rod and the calculator displays 3.2456 cm. The correct way to report this mean diameter is:
Show answer and why every option is right or wrong
Answer: A. A is correct — the reported value may not be finer than the least count of 0.01 cm, so 3.2456 cm rounds to 3.25 cm (NCERT Physics Lab Manual Class 11, Part 2, page 26).
Why B is wrong: B is wrong because the third decimal place claims a resolution of 0.001 cm, ten times finer than the instrument provides.
Why C is wrong: C is wrong because copying the full calculator display reports four decimals from an instrument that resolves two (trap: reporting more digits than instrument resolution).
Why D is wrong: D is wrong because it discards a digit the instrument genuinely supplies; averaging does not coarsen the least count.
MCQ 7CalculationPractice
A vernier calliper has 10 vernier divisions and 1 MSD = 1 mm. With the jaws closed, the vernier zero lies to the left of the main-scale zero and the 8th vernier division coincides with a main-scale mark. The internal diameter of a tube then reads 1.5 cm on the main scale with the 4th vernier division coinciding. The corrected internal diameter is:
Show answer and why every option is right or wrong
Answer: D. D is correct — LC = 1/10 mm = 0.01 cm, the negative zero error is −(10 − 8) × 0.01 = −0.02 cm, and the observed 1.5 + 4 × 0.01 = 1.54 cm is corrected as 1.54 − (−0.02) = 1.56 cm (NCERT Physics Lab Manual Class 11, Part 2, page 26).
Why A is wrong: A is wrong because it subtracts 0.02 cm; a negative zero error means the instrument under-reads, so the correction is added back (trap: zero-error sign confusion).
Why B is wrong: B is wrong because 1.54 cm is the uncorrected observed reading.
Why C is wrong: C is wrong because it keeps only the main-scale reading, 1.5 cm, and drops both the vernier contribution and the zero correction.
MCQ 8CalculationPractice
A vernier calliper has 1 MSD = 1 mm with 50 vernier divisions spanning 49 main-scale divisions. On closing the jaws, the vernier zero lies to the left of the main-scale zero and the 45th vernier division coincides. The depth strip then gives a main-scale reading of 3.4 cm with the 20th vernier division coinciding. The corrected depth of the vessel is:
Show answer and why every option is right or wrong
Answer: B. B is correct — LC = 1/50 mm = 0.002 cm, the negative zero error is −(50 − 45) × 0.002 = −0.010 cm, and the observed 3.4 + 20 × 0.002 = 3.440 cm becomes 3.440 + 0.010 = 3.450 cm (NCERT Physics Lab Manual Class 11, Part 2, page 26).
Why A is wrong: A is wrong because it subtracts the 0.010 cm offset; with the vernier zero to the left the instrument reads low, so the offset is added back (trap: zero-error sign confusion).
Why C is wrong: C is wrong because 3.440 cm is the observed reading before any zero-error correction.
Why D is wrong: D is wrong because it ignores the vernier contribution 20 × 0.002 cm and the zero error together.
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Experiment 01 Vernier Calipers: quick recall before you leave
How do you solve a Experiment 01 Vernier Calipers question? A worked example
- 1
Given.
A vernier calliper with 1 MSD = 1 mm and 10 vernier divisions (exact counting integer). With the jaws closed, the vernier zero lies to the right of the main-scale zero and the 2nd vernier division coincides with a main-scale mark. Three readings of the internal diameter of a vessel, taken with the upper jaws: main scale 2.5 cm with the 7th, 8th and 6th vernier divisions coinciding in turn (3 trials — exact count).
- 2
Required.
The mean internal diameter of the vessel, corrected and reported to the instrument's resolution.
- 3
Concept.
A vernier resolves 1 MSD / N. Every reading carries the closed-jaw offset, which is subtracted when the vernier zero sits to the right of the main-scale zero. The mean may not be quoted finer than the least count.
- 4
Formula.
LC = 1 MSD / N; observed = MSR + (VC × LC); corrected = observed − zero error.
- 5
Substitution.
LC = 1 mm / 10 = 0.1 mm = 0.01 cm. Zero error = +2 × 0.01 = +0.02 cm. Observed readings: 2.5 + 7 × 0.01 = 2.57 cm; 2.5 + 8 × 0.01 = 2.58 cm; 2.5 + 6 × 0.01 = 2.56 cm.
- 6
Calculation.
Mean observed = (2.57 + 2.58 + 2.56) / 3 = 7.71 / 3 = 2.57 cm. Corrected mean = 2.57 − 0.02 = 2.55 cm. The 10 vernier divisions, the 3 trials and the division numbers 7, 8 and 6 are exact counting integers — they do not limit the significant figures; only the least count of 0.01 cm does.
- 7
Final answer.
Internal diameter = 2.55 cm, quoted to the second decimal place in centimetres because the least count is 0.01 cm. Writing 2.5500 cm or 2.5533 cm would overstate the instrument's resolution.
- 8
Common trap.
Two separate slips share this problem. Adding the +0.02 cm instead of subtracting it gives 2.59 cm — a sign flip that moves the answer by twice the error. Carrying the calculator's 2.5700000 through to the answer sheet claims digits the caliper never produced.
- 9
Similar NEET-style question.
A vernier calliper with 1 MSD = 1 mm and 20 vernier divisions shows, on closed jaws, the vernier zero to the left of the main-scale zero with the 16th division coinciding. A cylinder reads 4.2 cm on the main scale with the 9th vernier division coinciding. Find the corrected diameter. (LC = 0.05 mm = 0.005 cm; zero error = −(20 − 16) × 0.005 = −0.020 cm; observed = 4.2 + 9 × 0.005 = 4.245 cm; corrected = 4.245 + 0.020 = 4.265 cm.)
What to remember before solving Experiment 01 Vernier Calipers questions
Which Experiment 01 Vernier Calipers formulas do you need for NEET?
1 formula — click to collapse
Least count of vernier callipers
Smallest length resolvable by a vernier; for standard 1-mm MSD with 10 VSD, LC = 0.1 mm = 0.01 cm.
| Symbol | Quantity | SI Unit |
|---|---|---|
| LC | least count | cm or mm |
| N | vernier divisions | - |
Valid when
- Standard linear vernier calliper
- 1 MSD value known
Where do students lose marks on Experiment 01 Vernier Calipers?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
4 items — click to collapse
Category: Overthinking
Reporting a measured length to more decimal places than the least count permits, e.g. writing 2.345 cm when LC = 0.01 cm.
When it triggers
Calculator output gives 4-5 decimals while the instrument's least count limits resolution to 2-3.
How to avoid
Round the final answer to the precision of the least-precise input or the limiting least count, whichever is coarser.
Category: Sign Convention
When the jaws of a vernier/screw gauge close, the zero of the auxiliary scale may not coincide with the main-scale zero — this offset must be subtracted (positive zero error) or added back (negative zero error) from every reading.
When it triggers
Question gives a vernier/screw gauge reading along with a 'zero error' or 'when the jaws are closed the reading is...'.
How to avoid
Subtract positive zero error, add back negative zero error, BEFORE recording the corrected length.
Root cause: concept gap
Correction
Always test for zero error by closing the jaws first; positive zero error is subtracted, negative is added back.
Root cause: concept gap
Correction
Final reading precision = least count of the limiting instrument; do NOT carry extra digits from a calculator.
More in Experimental Skills: 2 exam traps and mistakes · 2 formulas · 1 question pattern from its other lessons.
Experiment 01 Vernier Calipers questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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