Observation table: with main scale reading M and N the coinciding vernier division, the vernier scale reading is V = N × VC and the measured dimension is M + V; Corrected diameter = Mean observed diameter - Zero Error.
-- NCERT Physics Lab Manual Class 11, Part 2, p. 28Least Count
8 MCQs9-step worked example
Source: NCERT Physics and MeasurementPYQ coverage: NEET 2020, 2021, 2025Official key: NTA-verifiedLast updated: 7 Oct 2026
Try this first
The precision of a measurement made with an instrument depends mainly on which property of the instrument?
- A.Its measuring range.
- B.Its zero error.
- C.Its least count.
- D.Its overall size.
Tap to see the answer
Answer: C. C is correct: NCERT states that the precision of a measurement depends on the least count of the measuring instrument (NCERT Class 11 Physics, Chapter 1, page 3).
A is wrong: A is wrong because the range sets the largest value that can be read, not how finely a reading is resolved.
B is wrong: B is wrong because a zero error shifts every reading by the same amount; it is a correction, not the resolution of the instrument.
D is wrong: D is wrong because the physical size of the instrument has no bearing on the smallest division it can resolve.
Least Count, explained for NEET
The trap is the direction of the ratio. The least count is the smallest reading the instrument can resolve, so it is always smaller than the pitch of a screw gauge and smaller than one main-scale division of a vernier. If your answer for a least count comes out bigger than the pitch, the ratio was inverted.
NCERT Class 11 Physics, Chapter 1, page 3 says that the precision of a measurement depends on the least count of the measuring instrument.
Screw gauge: the least count is the pitch divided by the number of divisions on the circular scale. A pitch of 1 mm with 100 divisions gives 0.01 mm, and a pitch of 0.5 mm with 50 divisions also gives 0.01 mm. The total reading is the linear scale reading plus the circular scale reading times the least count (NCERT Physics Lab Manual Class 11, Part 2, page 35).
Vernier callipers: with 1 main scale division (MSD) = 1 mm and 10 vernier scale divisions (VSD) equal to 9 MSD, the vernier constant is 1 MSD - 1 VSD = 0.1 MSD = 0.1 mm = 0.01 cm (NCERT Physics Lab Manual Class 11, Part 2, page 26). The second trap sits here: when (N + 1) vernier divisions cover the length of N main divisions, one VSD is the shorter one, and the vernier constant is 1 MSD / (N + 1), not 1 MSD / N.
Zero error: if the vernier zero lies to the right of the main-scale zero when the jaws touch, the zero error is positive, readings are too large, and True reading = Observed reading - (+ zero error) (NCERT Physics Lab Manual Class 11, Part 2, page 27). For a vernier reading with main scale reading M and coinciding vernier division N, V = N x vernier constant, the measured value is M + V, and the corrected value subtracts the zero error (NCERT Physics Lab Manual Class 11, Part 2, page 28).
Watch-out: write the answer to the same decimal place as the least count, and subtract the zero error last.
How do you solve a Least Count question? A worked example
- 1
Given
A screw gauge has a least count of 0.002 mm and 250 divisions on its circular scale.
- 2
Required
The pitch of the screw gauge.
- 3
Concept
The least count of a screw gauge is the pitch divided by the number of circular-scale divisions, so the pitch is the least count times the number of divisions (NCERT Physics Lab Manual Class 11, Part 2, page 35).
- 4
Formula
least count = pitch / number of divisions, so pitch = least count x number of divisions.
- 5
Substitution
pitch = 0.002 mm x 250.
- 6
Calculation
pitch = 0.5 mm. The number of divisions is a count and is exact, so it does not affect the significant figures.
- 7
Final answer
The pitch is 0.5 mm.
- 8
Common trap
Dividing instead of multiplying: 0.002 / 250 gives a number far smaller than the least count, which cannot be a pitch. The pitch must be larger than the least count by the factor of the number of divisions.
- 9
Similar NEET-style question
A screw gauge has a least count of 0.01 mm and 40 divisions on its circular scale. What is its pitch? (Answer: pitch = least count × number of divisions = 0.01 × 40 = 0.4 mm.)
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Can you answer these Least Count 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 precision of a measurement made with an instrument depends mainly on which property of the instrument?
Show answer and why every option is right or wrong
Answer: C. C is correct: NCERT states that the precision of a measurement depends on the least count of the measuring instrument (NCERT Class 11 Physics, Chapter 1, page 3).
Why A is wrong: A is wrong because the range sets the largest value that can be read, not how finely a reading is resolved.
Why B is wrong: B is wrong because a zero error shifts every reading by the same amount; it is a correction, not the resolution of the instrument.
Why D is wrong: D is wrong because the physical size of the instrument has no bearing on the smallest division it can resolve.
MCQ 2Easy RecallPractice
Which expression gives the least count of a screw gauge?
Show answer and why every option is right or wrong
Answer: A. A is correct: the least count of a screw gauge is the pitch divided by the number of circular-scale divisions (NCERT Physics Lab Manual Class 11, Part 2, page 35).
Why B is wrong: B is wrong because multiplying by the number of divisions gives a length larger than the pitch; the least count must be smaller than the pitch. This is the pitch versus least count inversion.
Why C is wrong: C is wrong because it inverts the ratio and has the wrong unit; the pitch is the numerator.
Why D is wrong: D is wrong because it adds a length and a pure number, which has no meaning.
MCQ 3Easy RecallPractice
When the jaws of a vernier callipers are closed, the zero of the vernier scale lies to the right of the zero of the main scale. How is this zero error described, and how is the true reading obtained?
Show answer and why every option is right or wrong
Answer: B. B is correct: the vernier zero to the right of the main-scale zero is a positive zero error, readings are more than the actual value, and True reading = Observed reading - (+ zero error) (NCERT Physics Lab Manual Class 11, Part 2, page 27).
Why A is wrong: A is wrong because a zero to the right is positive, not negative, and adding the zero error would make the reading even larger.
Why C is wrong: C is wrong because it names the error correctly but adds it; a positive zero error makes readings too large, so it must be subtracted.
Why D is wrong: D is wrong because the error is positive when the vernier zero is to the right of the main-scale zero.
MCQ 4Direct ApplicationPractice
A screw gauge has a pitch of 1 mm and 100 divisions on its circular scale. What is its least count?
Show answer and why every option is right or wrong
Answer: D. D is correct: least count = pitch / number of divisions = 1 mm / 100 = 0.01 mm (NCERT Physics Lab Manual Class 11, Part 2, page 35).
Why A is wrong: A is wrong because 1 mm is the pitch itself; the least count is the pitch divided by the number of divisions.
Why B is wrong: B is wrong because 0.1 mm divides by 10 instead of 100; the circular scale has 100 divisions.
Why C is wrong: C is wrong because 0.001 mm divides by 1000, an extra factor of 10.
MCQ 5Direct ApplicationPractice
A screw gauge has a least count of 0.01 mm and 50 divisions on its circular scale. What is its pitch?
Show answer and why every option is right or wrong
Answer: A. A is correct: least count = pitch / number of divisions, so pitch = 0.01 mm x 50 = 0.5 mm (NCERT Physics Lab Manual Class 11, Part 2, page 35).
Why B is wrong: B is wrong because it divides the least count by 50, which swaps the roles of pitch and least count.
Why C is wrong: C is wrong because 0.01 x 50 is 0.5, not 5; the decimal point is out by a factor of 10.
Why D is wrong: D is wrong because 0.25 mm halves the correct value; nothing in the data introduces a factor of 1/2.
MCQ 6Direct ApplicationPractice
In a vernier callipers, 20 vernier scale divisions coincide with 19 main scale divisions, and 1 main scale division is 1 mm. What is the vernier constant (least count)?
Show answer and why every option is right or wrong
Answer: B. B is correct: 1 VSD = 19/20 MSD = 0.95 mm, so the vernier constant is 1 MSD - 1 VSD = 1 mm - 0.95 mm = 0.05 mm, which equals 1 MSD / 20 (NCERT Physics Lab Manual Class 11, Part 2, page 26).
Why A is wrong: A is wrong because 0.95 mm is the length of one vernier division, not the difference between a main and a vernier division.
Why C is wrong: C is wrong because 0.5 mm is out by a factor of 10; 1 mm / 20 is 0.05 mm.
Why D is wrong: D is wrong because 0.1 mm is the vernier constant for 10 divisions matching 9 main divisions, not for 20 matching 19.
MCQ 7CalculationPractice
For a vernier callipers, 10 VSD = 9 MSD and 1 MSD = 1 mm. With the jaws closed, the 3rd vernier division coincides with a main-scale division and the vernier zero is to the right of the main-scale zero. A rod is measured: the main scale reading is 4.2 cm and the 6th vernier division coincides. What is the corrected length?
Show answer and why every option is right or wrong
Answer: D. D is correct: vernier constant = 0.1 mm = 0.01 cm. Zero error = +3 x 0.01 = +0.03 cm. Observed = 4.2 + 6 x 0.01 = 4.26 cm. Corrected = 4.26 - 0.03 = 4.23 cm (NCERT Physics Lab Manual Class 11, Part 2, pages 27 and 28).
Why A is wrong: A is wrong because it adds the positive zero error (4.26 + 0.03) instead of subtracting it.
Why B is wrong: B is wrong because 4.26 cm is the observed reading; the zero error has not been subtracted.
Why C is wrong: C is wrong because 4.20 cm is the main scale reading alone; the vernier part 6 x 0.01 cm is missing.
MCQ 8CalculationPractice
A screw gauge has a pitch of 0.5 mm and 50 divisions on the circular scale, and no zero error. The linear scale reading is 3.5 mm and the 27th circular division is in line with the reference line. What is the reading?
Show answer and why every option is right or wrong
Answer: C. C is correct: least count = 0.5 mm / 50 = 0.01 mm. Total reading = linear scale reading + circular scale reading x least count = 3.5 + 27 x 0.01 = 3.77 mm (NCERT Physics Lab Manual Class 11, Part 2, page 35).
Why A is wrong: A is wrong because 3.5 + 27 x 0.1 = 6.2 mm uses a least count ten times too large.
Why B is wrong: B is wrong because 3.5 + 27 x 0.5 = 17.0 mm multiplies by the pitch instead of the least count.
Why D is wrong: D is wrong because 3.5 mm leaves out the circular scale reading, 27 x 0.01 mm.
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What to remember before solving Least Count questions
5 NCERT lines
Positive zero error: when the jaws touch, the zero of the vernier scale is shifted to the right of zero of the main scale, so readings are more than the actual value and True Reading = Observed reading - (+ Zero error).
-- NCERT Physics Lab Manual Class 11, Part 2, p. 27Key Fact
Vernier callipers: 10 vernier divisions = 9 main scale divisions gives vernier constant 0.01 cm
With 1 main scale division (MSD) = 1 mm = 0.1 cm and 10 vernier scale divisions = 9 main scale divisions, the vernier constant is 1 MSD - 1 VSD = 0.1 MSD, so VC = 0.1 mm = 0.01 cm.
-- NCERT Physics Lab Manual Class 11, Part 2, p. 26Least count of a screw gauge = pitch / number of divisions on the circular scale: a pitch of 1 mm with 100 divisions gives 1 mm/100 = 0.01 mm, and a pitch of 0.5 mm with 50 divisions gives 0.5 mm/50 = 0.01 mm. Total reading = linear scale reading + circular scale reading × least count.
-- NCERT Physics Lab Manual Class 11, Part 2, p. 35Significant figures indicate, as already mentioned, the precision of measurement which depends on the least count of the measuring instrument.
-- NCERT Class 11 Physics, Ch. 1, p. 3Where do students lose marks on Least Count?
2 traps and mistakes
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Similar Terms
Student swaps which is the input vs the output: least count = pitch / N (where N is the number of circular-scale divisions). Distractors offer the ratio inverted or the wrong unit.
When it triggers
Question gives one of (pitch, N, least count) and asks for another; distractors offer the inverted ratio or off-by-factor-of-10.
How to avoid
Anchor on the definition: least count is the SMALLEST measurement the instrument can resolve. It is always SMALLER than the pitch. So pitch = LC × N (and not LC = pitch × N).
Category: Similar Terms
Confusing whether N or N+1 is the smaller count when (N+1) divisions of vernier match N divisions of main scale.
When it triggers
Question gives '(N+1) divisions of vernier coincide with N divisions of main' or similar phrasing.
How to avoid
Always interpret carefully: N+1 vernier divisions span the SAME LENGTH as N main divisions. So 1 VSD = (N/(N+1)) MSD; vernier constant = 1 MSD - 1 VSD = 1 MSD / (N+1). Result smaller than 1 MSD.
Least Count: NEET previous year questions (PYQs) with answers
3 questions from NEET 2020, 2021, 2025, answers verified against NTA official keys
NEET 2025
15.00 cm
25.18 cm
35.08 cm
44.98 cm
NTA Answer: Option 4(final)
Why the other options are wrong
- Option 1: 5.00 cm is the main-scale reading alone. The vernier adds 8 × 0.01 cm ('Vernier constant (VC ) = 0.1 mm = 0.01 cm'), giving 5.08 cm, and then the zero error has to be removed: 5.08 − 0.1 = 4.98 cm.
- Option 2: 5.18 cm adds the zero error instead of subtracting it. With the vernier zero 'shifted to the right of zero of the main scale', readings are too large, and 'Corrected diameter = Mean observed diameter – Zero Error' = 5.08 − 0.1 = 4.98 cm.
- Option 3: 5.08 cm is the observed reading, M + N × VC = 5 + 8 × 0.01 cm, before the zero error is removed. 'Corrected diameter = Mean observed diameter – Zero Error' = 5.08 − 0.10 = 4.98 cm.
NEET 2021
10.052 cm
20.52 cm
30.026 cm
40.26 cm
NTA Answer: Option 1(final)
Why the other options are wrong
- Option 2: 0.52 cm has the right digits in the wrong unit. 'Total reading = linear scale reading + circular scale reading × least count' = 0 + 52 × 0.01 mm = 0.52 mm, which is 0.052 cm.
- Option 3: 0.026 cm halves the reading, as if the least count were 0.005 mm. With 1 mm per 100 divisions the least count is 0.01 mm (the page’s example is 'a pitch of 1mm and 100 divisions on the circular scale'), so the diameter is 52 × 0.01 mm = 0.52 mm = 0.052 cm.
- Option 4: 0.26 cm is both halved and ten times too large. 'Total reading = linear scale reading + circular scale reading × least count' = 0 + 52 × 0.01 mm = 0.52 mm = 0.052 cm.
NEET 2020
11.0 mm
20.01 mm
30.25 mm
40.5 mm
NTA Answer: Option 4(final)
Why the other options are wrong
- Option 1: A 1.0 mm pitch with 50 divisions gives a least count of 1/50 = 0.02 mm, not 0.01 mm. Least count = pitch / number of circular-scale divisions (the page’s example is 'a pitch of 1mm and 100 divisions on the circular scale', giving 0.01 mm), so here pitch = 0.01 mm × 50 = 0.5 mm.
- Option 2: 0.01 mm is the least count itself, the distance moved for one circular-scale division, which 'is called the least count of the instrument'. The pitch is the distance moved in one full rotation of 50 divisions: 50 × 0.01 mm = 0.5 mm.
- Option 3: A 0.25 mm pitch with 50 divisions gives a least count of 0.25/50 = 0.005 mm, half the stated value. The least count, 'the least distance that can be measured accurately by the instrument', is pitch / divisions, so pitch = 0.01 mm × 50 = 0.5 mm.
How does NEET ask about Least Count?
2 recurring patterns from past papers
Given the least count and the number of circular-scale divisions of a screw gauge, find the pitch (or vice versa). Formula: least count = pitch / (number of circular scale divisions). Common shape: LC = 0.01 mm, 50 divisions; find pitch (= 0.5 mm).
Direct ApplicationEasy
Common distractors
swaps pitch and LC
Confusing which side of the formula is asked
Vernier calipers: (N+1) divisions of vernier scale coincide with N divisions of main scale; given main-scale division (1 MSD), find vernier constant (least count). Formula: VC = 1 MSD - 1 VSD = 1 MSD × (1 - N/(N+1)) = 1 MSD / (N+1). Common shape: 1 MSD = 0.1 mm, k VSD = (k+1) MSD; find VC.
Direct ApplicationEasy
Common distractors
swap N and N+1
Confusing which side has the larger count
More in Units and Measurements: 8 exam traps and mistakes · 5 formulas · 7 question patterns from its other lessons.
Sources
NCERT refs: Class 11 Physics Chapter 1, p.3
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.
Every past-paper question from this chapter:
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