Vernier Calliper — JEE Main Physics MCQs with Solutions
Free JEE Main 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
C. 1 main scale division − 1 vernier scale division ✓ Correct
D. 1 vernier division only
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. 0.01 mm
B. 1 mm
C. 0.9 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. Measure temperature
B. Replace the main scale
C. Measure fractions of the smallest main-scale division ✓ Correct
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 only
B. Main scale reading + (coinciding vernier division × least count) ✓ Correct
C. Vernier reading only
D. Main scale reading × least count
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. 2.14 cm ✓ Correct
B. 2.10 cm
C. 2.104 cm
D. 2.4 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. 3.007 cm
B. 3.07 cm ✓ Correct
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. Exactly on the zero
C. Off the scale
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. Ignoring it
C. Adding the zero error
D. Subtracting the zero error from the observed reading ✓ Correct
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. Adding the magnitude of the zero error to the observed reading ✓ Correct
B. Subtracting it
C. Dividing by it
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. +0.6 cm
C. +0.06 cm ✓ Correct
D. −0.06 cm
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. millimetres ✓ Correct
B. inches only
C. centimetres only
D. micrometres
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. 1 mm
D. 0.05 mm ✓ Correct
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. Temperature
C. Only depth
D. Only internal dimensions
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. Depth of a beaker
C. Mass
D. External diameter
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 thin depth rod at the end ✓ Correct
B. The upper jaws
C. The lower jaws
D. The main scale only
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. 1.95 cm
B. 1.509 cm
C. 1.59 cm ✓ Correct
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. 0.1 mm (0.01 cm) ✓ Correct
B. 0.01 mm
C. 1 mm
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. 2.33 cm
B. 2.30 cm
C. 2.35 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. 4.21 cm
B. 4.03 cm
C. 4.15 cm ✓ Correct
D. 4.18 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. Main scale reading
B. Zero error
C. Least count ✓ Correct
D. Pitch
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−1) MSD
C. (1 MSD)×N
D. (N) MSD
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. Take the vernier line that most nearly coincides ✓ Correct
B. Take the last line
C. Discard the reading
D. Take the first 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. 5.28 cm
B. 5.03 cm
C. 5.31 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 before the vernier zero ✓ Correct
C. Just after the vernier zero
D. Where the vernier ends
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. Higher precision ✓ Correct
B. Larger zero error
C. Lower precision
D. Bigger main scale
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. 0.02 mm ✓ Correct
B. 0.05 mm
C. 0.5 mm
D. 0.1 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. A systematic error to be corrected in every reading ✓ Correct
B. Never correctable
C. The least count
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. 6.2 cm
B. 6.002 cm
C. 6.02 cm ✓ Correct
D. 6.20 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 time
C. Measuring temperature
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. 0.1 mm
B. 0.5 mm
C. 0.02 mm ✓ Correct
D. 0.05 mm
Solution: LC = 1 MSD / N = 0.5 mm / 25 = 0.02 mm.