Screw gauge measures small thicknesses (sheet, wire) to 0.001 cm. Pitch = linear distance moved per full rotation; least count = pitch / number of head-scale divisions. Zero error (when zero of head scale does not coincide with reference line) is added/subtracted. Backlash error is avoided by rotating in one direction only when taking final reading.
-- NCERT Physics Lab Manual Class 11, Part 2, p. 35Experiment 02 Screw Gauge
8 MCQs2 revision cards9-step worked example
Source: NCERT Experimental SkillsOfficial key: NTA-verifiedLast updated: 24 Sep 2026
Experiment 02 Screw Gauge, explained for NEET
The number a screw gauge shows is not the number you report. Close the studs on nothing at all and the circular scale usually refuses to sit on zero — that offset rides on every reading you take afterwards, and a common confusion in this experiment is losing its sign.
Fix the vocabulary first. The pitch is how far the screw tip advances along the axis in one full turn of the thimble. The least count is pitch divided by the number of divisions on the circular scale — for the standard instrument, 0.5 mm over 50 divisions gives 0.01 mm. Pitch and least count are different quantities; an option list that offers one where the stem asks for the other is the cheapest mark in this topic.
Now the offset. With the stud and screw face in contact, look at where the circular scale's zero sits relative to the reference line. If the zero has gone below the line, the instrument is already reading a little too much: that is a positive zero error, and you subtract it. If the zero sits above the line, the instrument reads short: a negative zero error, which you add back. Correct first, then record. The NCERT Physics Laboratory Manual sets out the closed-stud check as a step before any observation (NCERT Physics Lab Manual Class 11, Part 2, page 33).
NEET asks this as a two-step arithmetic item: build the total from main scale plus circular-scale count × least count, then apply the correction. The negative-error case is where marks leak, because "add back" feels wrong when the word is error.
Watch-out: the ratchet exists so every object is squeezed with the same pressure. Turn the thimble directly and you compress a soft wire into a smaller "diameter" than it has.
Can you answer these Experiment 02 Screw Gauge 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
In a screw gauge, the pitch is defined as:
Show answer and why every option is right or wrong
Answer: C. C is correct — pitch is the axial advance per full turn of the thimble, the quantity that goes in the numerator of the least-count relation (NCERT Physics Lab Manual Class 11, Part 2, page 33).
Why A is wrong: A is wrong because the circular-division count is the denominator of the least-count relation, not the pitch itself; swapping the two is the standard pitch/least-count mix-up.
Why B is wrong: B is wrong because the screw's total travel is the instrument's range, which has no role in fixing either the pitch or the least count.
Why D is wrong: D is wrong because the smallest resolvable length is the least count, which equals pitch divided by the circular-scale divisions — a derived quantity, not the pitch.
MCQ 2Easy RecallPractice
The ratchet fitted at the outer end of the thimble of a screw gauge is provided in order to:
Show answer and why every option is right or wrong
Answer: D. D is correct — the ratchet slips once a set torque is reached, so the object is gripped with a reproducible force and is not crushed (NCERT Physics Lab Manual Class 11, Part 2, page 33).
Why A is wrong: A is wrong because the zero error survives the ratchet completely; it is found by closing the studs and is removed only by arithmetic on each reading.
Why B is wrong: B is wrong because the circular-scale division count is fixed when the instrument is made and no attachment alters it.
Why C is wrong: C is wrong because locking is the job of the separate lock nut; the ratchet controls force, not position.
MCQ 3Easy RecallPractice
A screw gauge is found to have a positive zero error. To obtain the corrected reading, that zero error must be:
Show answer and why every option is right or wrong
Answer: A. A is correct — a positive zero error means the instrument already reads high with nothing between the studs, so the offset is subtracted from each observed reading (NCERT Physics Lab Manual Class 11, Part 2, page 33).
Why B is wrong: B is wrong because adding is the treatment for a negative zero error; applying it to a positive error doubles the offset instead of removing it.
Why C is wrong: C is wrong because the offset is mechanical and present in every single reading, so every reading needs the same correction.
Why D is wrong: D is wrong because a constant offset does not cancel in a mean — averaging biased readings gives a biased average, which is exactly the habit of reporting the raw reading.
MCQ 4Direct ApplicationPractice
A screw gauge has a pitch of 0.5 mm and 50 divisions (exact) on its circular scale. Its least count is:
Show answer and why every option is right or wrong
Answer: B. B is correct — least count = pitch / number of circular divisions = 0.5 mm / 50 = 0.01 mm, the standard value for this instrument (NCERT Physics Lab Manual Class 11, Part 2, page 33).
Why A is wrong: A is wrong because 0.001 mm comes from dividing by 500 rather than 50, one decimal place too far.
Why C is wrong: C is wrong because 0.05 mm is pitch divided by 10; it uses a vernier-style division count instead of the 50 divisions stated in the stem.
Why D is wrong: D is wrong because 0.1 mm is obtained by dividing the 50 divisions into the pitch upside-down, or by quoting a vernier least count from memory.
MCQ 5Direct ApplicationPractice
When the thimble of a screw gauge is given 6 complete rotations (exact count), the screw tip advances 3.0 mm along the main scale. The pitch of the screw is:
Show answer and why every option is right or wrong
Answer: D. D is correct — pitch is the advance per single rotation, so pitch = 3.0 mm / 6 = 0.50 mm; the rotation count is an exact integer and does not limit the significant figures (NCERT Physics Lab Manual Class 11, Part 2, page 33).
Why A is wrong: A is wrong because 0.05 mm would be the pitch divided again by the circular divisions; that quotient is a least count, not a pitch.
Why B is wrong: B is wrong because 0.30 mm comes from dividing by 10 instead of by the 6 rotations actually performed.
Why C is wrong: C is wrong because 0.60 mm divides the advance by 5 rotations; the stem specifies 6.
MCQ 6Direct ApplicationPractice
With nothing between the studs of a screw gauge of least count 0.01 mm, the circular-scale zero lies below the reference line and the 4th division coincides with it. A thin sheet then gives an observed reading of 2.56 mm. The corrected thickness of the sheet is:
Show answer and why every option is right or wrong
Answer: A. A is correct — the zero error is +4 × 0.01 mm = +0.04 mm, and a positive zero error is subtracted: 2.56 mm − 0.04 mm = 2.52 mm (NCERT Physics Lab Manual Class 11, Part 2, page 33).
Why B is wrong: B is wrong because 2.60 mm adds the offset instead of subtracting it — the sign flip that the zero-error trap is built on.
Why C is wrong: C is wrong because 2.56 mm is the raw observed reading, reported without applying the closed-stud correction at all.
Why D is wrong: D is wrong because 2.48 mm subtracts the offset twice, treating 0.04 mm as if it had to be removed from both scales.
MCQ 7CalculationPractice
A screw gauge of pitch 0.5 mm carries 50 divisions (exact) on its circular scale. With the studs closed, the circular-scale zero lies above the reference line and the 45th division coincides with it. A wire then gives a main-scale reading of 2.5 mm with the 28th circular division against the reference line. The corrected diameter of the wire is:
Show answer and why every option is right or wrong
Answer: C. C is correct — least count = 0.01 mm, observed reading = 2.5 + 28 × 0.01 = 2.78 mm, and the negative zero error of −(50 − 45) × 0.01 = −0.05 mm is added back, giving 2.83 mm (NCERT Physics Lab Manual Class 11, Part 2, page 33).
Why A is wrong: A is wrong because 2.73 mm subtracts 0.05 mm; subtraction is the treatment for a positive zero error, and here the zero sits above the line.
Why B is wrong: B is wrong because 2.78 mm is the observed reading left uncorrected, the raw-reading habit this experiment is designed to break.
Why D is wrong: D is wrong because 3.23 mm reads the closed-stud coincidence as 45 × 0.01 = 0.45 mm of error instead of counting the 5 divisions short of a full turn.
MCQ 8Concept TrapPractice
Which single change to a screw gauge would halve its least count?
Show answer and why every option is right or wrong
Answer: B. B is correct — least count = pitch / circular divisions, so doubling the denominator alone halves it (NCERT Physics Lab Manual Class 11, Part 2, page 33).
Why A is wrong: A is wrong because pitch sits in the numerator; doubling it doubles the least count, making the instrument coarser rather than finer.
Why C is wrong: C is wrong because halving the denominator doubles the least count — the same coarsening, reached from the other side of the fraction.
Why D is wrong: D is wrong because doubling numerator and denominator together leaves the ratio, and therefore the least count, exactly as it was.
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Experiment 02 Screw Gauge: quick recall before you leave
How do you solve a Experiment 02 Screw Gauge question? A worked example
- 1
Given
Pitch of the screw = 0.50 mm. Number of circular-scale divisions = 50 (exact count). With the stud and screw face in contact, the circular-scale zero lies below the reference line and the 3rd division (exact count) coincides with it. On a wire: main-scale reading = 1.5 mm, circular-scale division against the reference line = 42 (exact count).
- 2
Required
The corrected diameter of the wire, reported to the precision the instrument actually supports.
- 3
Concept
A screw gauge reading is built in two parts — whole pitches read off the main scale, plus a fraction of a pitch read off the circular scale — and then corrected for the offset the instrument shows with nothing between its faces. The zero below the reference line means the instrument reads high, so the offset is positive and comes off the total.
- 4
Formula
LC = pitch / (circular divisions); observed reading = main-scale reading + (circular-scale division × LC); corrected reading = observed reading − zero error.
- 5
Substitution
LC = 0.50 mm / 50. Zero error = +3 × LC. Observed reading = 1.5 mm + 42 × LC.
- 6
Calculation
LC = 0.010 mm. Zero error = +3 × 0.010 = +0.030 mm. Observed reading = 1.5 + 0.420 = 1.920 mm. Corrected reading = 1.920 − 0.030 = 1.890 mm. The division counts (50, 3 and 42) are exact counting integers, so they place no limit on the significant figures; the precision of the answer is set entirely by the least count of 0.010 mm.
- 7
Final answer
Diameter of the wire = 1.89 mm — two decimal places in millimetres, matching the 0.01 mm least count.
- 8
Common trap
Reporting 1.92 mm (the observed reading, never corrected) or 1.95 mm (the offset added instead of subtracted). Decide the sign by looking at where the circular-scale zero sits with the studs closed — below the line means positive, and positive comes off.
- 9
Similar NEET-style question
A screw gauge of pitch 1.0 mm and 100 circular divisions (exact) shows, with the studs closed, its circular-scale zero above the reference line with the 97th division coinciding. A sheet gives a main-scale reading of 4.0 mm with the 12th circular division against the line. Find the corrected thickness. (Set up the least count first, then decide the sign of the offset before touching the arithmetic.)
What to remember before solving Experiment 02 Screw Gauge questions
Which Experiment 02 Screw Gauge formulas do you need for NEET?
1 formula — click to collapse
Least count of screw gauge
Smallest length resolvable by a screw gauge; typical pitch 0.5 mm, 50 head divisions ⇒ LC = 0.01 mm.
| Symbol | Quantity | SI Unit |
|---|---|---|
| p | pitch | mm |
| n_circ | circular divisions | - |
Valid when
- Standard micrometer screw gauge
- Backlash error eliminated
Where do students lose marks on Experiment 02 Screw Gauge?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
2 items — click to collapse
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.
More in Experimental Skills: 4 exam traps and mistakes · 2 formulas · 1 question pattern from its other lessons.
Experiment 02 Screw Gauge 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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