Vernier Calliper — NEET Physics MCQs with Solutions
Free NEET Physics Vernier Calliper MCQs with step-by-step solutions (30 questions). Part of Units and Dimensions. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Vernier Calliper · medium · theory
The least count of a vernier calliper is defined as
A. Pitch ÷ divisions
B. 1 main scale division − 1 vernier scale division ✓ Correct
C. 1 vernier division only
D. 1 main scale division
Solution: Least count = value of smallest main scale division − value of one vernier division = (1 MSD) − (1 VSD).
Q2 — Vernier Calliper · medium · numerical
In a vernier calliper, 10 vernier divisions coincide with 9 main scale divisions. If 1 MSD = 1 mm, the least count is
A. 1 mm
B. 0.9 mm
C. 0.01 mm
D. 0.1 mm ✓ Correct
Solution: 1 VSD = 0.9 mm, so LC = 1 − 0.9 = 0.1 mm = 0.01 cm.
Q3 — Vernier Calliper · easy · theory
The vernier scale is used to
A. Replace the main scale
B. Measure fractions of the smallest main-scale division ✓ Correct
C. Measure temperature
D. Measure very large distances
Solution: The vernier lets you read a fraction of the main-scale division, improving precision.
Q4 — Vernier Calliper · medium · theory
The reading of a vernier calliper is
A. Main scale reading + (coinciding vernier division × least count) ✓ Correct
B. Main scale reading only
C. Main scale reading × least count
D. Vernier reading only
Solution: Total reading = MSR + (VSD that coincides) × LC.
Q5 — Vernier Calliper · medium · numerical
A vernier calliper has main scale reading 2.1 cm and the 4th vernier division coincides. If LC = 0.01 cm, the reading is
A. 2.4 cm
B. 2.104 cm
C. 2.14 cm ✓ Correct
D. 2.10 cm
Solution: 2.1 + 4×0.01 = 2.1 + 0.04 = 2.14 cm.
Q6 — Vernier Calliper · medium · numerical
A vernier calliper has MSR = 3.0 cm and the 7th vernier line coincides (LC = 0.01 cm). The measured length is
A. 3.07 cm ✓ Correct
B. 3.007 cm
C. 3.70 cm
D. 3.7 cm
Solution: 3.0 + 7×0.01 = 3.07 cm.
Q7 — Vernier Calliper · medium · theory
A vernier calliper shows a positive zero error when, with the jaws closed, the vernier zero lies
A. To the right of the main scale zero ✓ Correct
B. Off the scale
C. Exactly on the zero
D. To the left of the main scale zero
Solution: If the vernier zero is ahead (right) of the main scale zero, the error is positive.
Q8 — Vernier Calliper · medium · theory
If a vernier calliper has a positive zero error, the correct reading is obtained by
A. Multiplying by it
B. Subtracting the zero error from the observed reading ✓ Correct
C. Adding the zero error
D. Ignoring it
Solution: Correct reading = observed reading − (positive) zero error.
Q9 — Vernier Calliper · medium · theory
If a vernier calliper has a negative zero error, the correct reading is obtained by
A. Dividing by it
B. Subtracting it
C. Adding the magnitude of the zero error to the observed reading ✓ Correct
D. Ignoring it
Solution: For negative zero error, the correction is added: correct = observed − (negative) = observed + |error|.
Q10 — Vernier Calliper · medium · numerical
With the jaws closed, the 6th vernier division coincides with a main scale mark (LC = 0.01 cm). The zero error is
A. 0
B. −0.06 cm
C. +0.6 cm
D. +0.06 cm ✓ Correct
Solution: Positive zero error = 6 × 0.01 = +0.06 cm; this must be subtracted from readings.
Q11 — Vernier Calliper · easy · theory
The main scale of a common vernier calliper is graduated in
A. centimetres only
B. micrometres
C. inches only
D. millimetres ✓ Correct
Solution: The main scale is typically marked in mm (with cm labels).
Q12 — Vernier Calliper · medium · numerical
If 20 vernier divisions coincide with 19 main scale divisions (1 MSD = 1 mm), the least count is
A. 0.5 mm
B. 0.1 mm
C. 0.05 mm ✓ Correct
D. 1 mm
Solution: 1 VSD = 19/20 = 0.95 mm; LC = 1 − 0.95 = 0.05 mm.
Q13 — Vernier Calliper · medium · theory
A vernier calliper reads a length using both jaws for
A. External (outside) dimensions like a rod's diameter ✓ Correct
B. Only depth
C. Only internal dimensions
D. Temperature
Solution: The lower (main) jaws measure external dimensions; upper jaws measure internal, and the depth rod measures depth.
Q14 — Vernier Calliper · easy · theory
The upper jaws of a vernier calliper are used to measure
A. Internal diameter of a hollow object ✓ Correct
B. External diameter
C. Depth of a beaker
D. Mass
Solution: The smaller upper jaws fit inside a hollow object to measure its internal diameter.
Q15 — Vernier Calliper · easy · theory
The depth of a beaker is measured with a vernier calliper using
A. The main scale only
B. The thin depth rod at the end ✓ Correct
C. The lower jaws
D. The upper jaws
Solution: The depth-measuring rod extends from the end of the vernier as the instrument opens.
Q16 — Vernier Calliper · medium · numerical
Least count = 0.01 cm. A reading gives MSR = 1.5 cm and vernier coincidence at division 9. The length is
A. 1.59 cm ✓ Correct
B. 1.509 cm
C. 1.95 cm
D. 1.9 cm
Solution: 1.5 + 9×0.01 = 1.59 cm.
Q17 — Vernier Calliper · easy · numerical
The precision of a vernier calliper is generally
A. 1 mm
B. 0.01 mm
C. 0.1 mm (0.01 cm) ✓ Correct
D. 1 cm
Solution: A standard vernier calliper resolves down to 0.1 mm = 0.01 cm.
Q18 — Vernier Calliper · medium · numerical
A vernier calliper with a negative zero error of −0.02 cm gives an observed reading of 2.35 cm. The corrected reading is
A. 2.35 cm
B. 2.30 cm
C. 2.33 cm
D. 2.37 cm ✓ Correct
Solution: Correct = observed − (−0.02) = 2.35 + 0.02 = 2.37 cm.
Q19 — Vernier Calliper · medium · numerical
A vernier calliper with a positive zero error of +0.03 cm gives an observed reading of 4.18 cm. The corrected reading is
A. 4.15 cm ✓ Correct
B. 4.03 cm
C. 4.18 cm
D. 4.21 cm
Solution: Correct = observed − (+0.03) = 4.18 − 0.03 = 4.15 cm.
Q20 — Vernier Calliper · medium · theory
The vernier constant is another name for the
A. Pitch
B. Main scale reading
C. Least count ✓ Correct
D. Zero error
Solution: The vernier constant IS the least count of the vernier calliper.
Q21 — Vernier Calliper · medium · theory
If N vernier divisions are equal to (N−1) main scale divisions, the least count is
A. (1 MSD)/N ✓ Correct
B. (N) MSD
C. (N−1) MSD
D. (1 MSD)×N
Solution: 1 VSD = (N−1)/N MSD, so LC = 1 MSD − 1 VSD = (1 MSD)/N.
Q22 — Vernier Calliper · medium · theory
When no vernier division exactly coincides, one should
A. Discard the reading
B. Take the first line
C. Take the vernier line that most nearly coincides ✓ Correct
D. Take the last line
Solution: The vernier division that best lines up with a main scale mark is used.
Q23 — Vernier Calliper · medium · numerical
A vernier calliper with LC 0.01 cm reads MSR = 0.0 cm and vernier coincidence at 3 with jaws touching. The true length of an object then reading 5.28 cm is
A. 5.31 cm
B. 5.03 cm
C. 5.28 cm
D. 5.25 cm ✓ Correct
Solution: Positive zero error = +0.03 cm; true = 5.28 − 0.03 = 5.25 cm.
Q24 — Vernier Calliper · easy · theory
The main scale reading is taken as the mark on the main scale that is
A. At the last main scale mark
B. Just after the vernier zero
C. Where the vernier ends
D. Just before the vernier zero ✓ Correct
Solution: MSR is read at the main-scale division immediately to the left of the vernier zero.
Q25 — Vernier Calliper · medium · theory
Smaller least count of a vernier means
A. Larger zero error
B. Lower precision
C. Bigger main scale
D. Higher precision ✓ Correct
Solution: A smaller least count lets you measure finer differences — greater precision.
Q26 — Vernier Calliper · medium · numerical
A vernier has 50 divisions coinciding with 49 MSD (1 MSD = 1 mm). Its least count is
A. 0.05 mm
B. 0.02 mm ✓ Correct
C. 0.1 mm
D. 0.5 mm
Solution: 1 VSD = 49/50 = 0.98 mm; LC = 1 − 0.98 = 0.02 mm.
Q27 — Vernier Calliper · medium · theory
The zero error of a vernier calliper is
A. The least count
B. Never correctable
C. A systematic error to be corrected in every reading ✓ Correct
D. A random error
Solution: Zero error is constant (systematic) and is subtracted (with sign) from every reading.
Q28 — Vernier Calliper · medium · numerical
A reading of 6.0 cm on the main scale with vernier coincidence at 2 (LC = 0.01 cm) gives
A. 6.02 cm ✓ Correct
B. 6.2 cm
C. 6.20 cm
D. 6.002 cm
Solution: 6.0 + 2×0.01 = 6.02 cm.
Q29 — Vernier Calliper · easy · theory
The vernier calliper was invented to overcome the limitation of
A. Weighing objects
B. Measuring temperature
C. Measuring time
D. Reading fractions smaller than the main-scale least division ✓ Correct
Solution: It allows measurement of lengths finer than one main-scale division.
Q30 — Vernier Calliper · medium · numerical
If 1 MSD = 0.5 mm and 25 vernier divisions equal 24 MSD, the least count is
A. 0.05 mm
B. 0.1 mm
C. 0.02 mm ✓ Correct
D. 0.5 mm
Solution: LC = 1 MSD / N = 0.5 mm / 25 = 0.02 mm.