Measuring Instruments — NEET Physics MCQs with Solutions
Free NEET Physics Measuring Instruments MCQs with step-by-step solutions (20 questions). Part of Units and Dimensions. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Measuring Instruments · easy · numerical
In a Vernier caliper, 1 MSD = 1 mm and 10 Vernier scale divisions coincide with 9 main scale divisions. The least count of the caliper is:
A. 0.05 cm
B. 0.001 cm
C. 0.1 cm
D. 0.01 cm ✓ Correct
Solution: 1 VSD = (9/10) MSD = 0.9 mm. LC = 1 MSD − 1 VSD = 1 mm − 0.9 mm = 0.1 mm = 0.01 cm.
Q2 — Measuring Instruments · easy · numerical
A Vernier caliper has 1 MSD = 0.5 mm. If 50 divisions on the Vernier scale coincide with 49 divisions on the main scale, its least count is:
A. 0.01 mm ✓ Correct
B. 0.05 mm
C. 0.005 mm
D. 0.02 mm
Solution: 50 VSD = 49 MSD, so LC = (1 MSD)/50 = 0.5 mm / 50 = 0.01 mm.
Q3 — Measuring Instruments · easy · numerical
In a specially designed Vernier caliper, 1 MSD = 1 mm and 20 divisions of the Vernier scale coincide with 16 main scale divisions. What is the least count of this instrument?
A. 0.05 mm
B. 0.2 mm ✓ Correct
C. 0.02 mm
D. 0.1 mm
Solution: 20 VSD = 16 MSD, so 1 VSD = 0.8 mm. LC = 1 mm − 0.8 mm = 0.2 mm.
Q4 — Measuring Instruments · easy · numerical
For a Vernier caliper with 1 MSD = 1 mm and 10 VSD = 9 MSD (zero error = 0), the zero of the Vernier lies between 3.2 cm and 3.3 cm on the main scale, and the 6th Vernier division coincides with a main scale division. The length of the rod is:
A. 3.36 cm
B. 3.26 cm ✓ Correct
C. 3.20 cm
D. 3.24 cm
Solution: LC = 0.01 cm. Length = MSR + (V × LC) = 3.2 + (6 × 0.01) = 3.26 cm.
Q5 — Measuring Instruments · easy · numerical
In a Vernier caliper (1 MSD = 1 mm, 10 VSD = 9 MSD), when the two jaws touch, the zero of the Vernier scale lies to the right of the main scale zero, and the 3rd Vernier division coincides with a main scale division. What is the zero error?
A. +0.07 cm
B. +0.03 cm ✓ Correct
C. −0.07 cm
D. −0.03 cm
Solution: Vernier zero to the right → positive zero error = +(n × LC) = +(3 × 0.01) = +0.03 cm.
Q6 — Measuring Instruments · medium · numerical
In a Vernier caliper with 10 VSD = 9 MSD (1 MSD = 1 mm), when the jaws touch, the zero of the Vernier scale is to the left of the main scale zero, and the 7th Vernier division coincides with a main scale division. The zero error is:
A. −0.07 cm
B. +0.07 cm
C. +0.03 cm
D. −0.03 cm ✓ Correct
Solution: Vernier zero to the left → negative zero error = −(N − n) × LC = −(10 − 7) × 0.01 = −0.03 cm.
Q7 — Measuring Instruments · medium · numerical
A Vernier caliper has 1 MSD = 1 mm and 10 VSD = 9 MSD. When the jaws close, the zero of the Vernier scale is to the right of the main scale zero and the 4th VSD coincides. While measuring a sphere, main scale reading is 2.5 cm and the 8th VSD coincides. The true diameter of the sphere is:
A. 2.62 cm
B. 2.54 cm ✓ Correct
C. 2.50 cm
D. 2.58 cm
Solution: Zero error = +(4 × 0.01) = +0.04 cm. Measured = 2.5 + (8 × 0.01) = 2.58 cm. True = 2.58 − 0.04 = 2.54 cm.
Q8 — Measuring Instruments · medium · numerical
A Vernier caliper (1 MSD = 1 mm, 10 VSD = 9 MSD) has a negative zero error. When the jaws touch, the 6th Vernier division coincides. While measuring the thickness of a plate, the main scale reads 1.8 cm and the 5th VSD coincides. The actual thickness is:
A. 1.89 cm ✓ Correct
B. 1.85 cm
C. 1.79 cm
D. 1.81 cm
Solution: Zero error = −(10 − 6) × 0.01 = −0.04 cm. Measured = 1.8 + (5 × 0.01) = 1.85 cm. True = 1.85 − (−0.04) = 1.89 cm.
Q9 — Measuring Instruments · medium · numerical
A Vernier scale has 20 divisions which coincide with 19 main scale divisions, and 1 MSD = 0.5 mm. When jaws are brought together, the Vernier zero is to the right of the main scale zero and the 6th VSD coincides. When measuring a cylinder inner diameter, MSR = 1.40 cm and the 14th VSD coincides. The true inner diameter is:
A. 1.425 cm
B. 1.415 cm
C. 1.435 cm
D. 1.420 cm ✓ Correct
Solution: LC = 0.5 mm / 20 = 0.025 mm = 0.0025 cm. Zero error = +(6 × 0.0025) = +0.015 cm. Measured = 1.40 + (14 × 0.0025) = 1.435 cm. True = 1.435 − 0.015 = 1.420 cm.
Q10 — Measuring Instruments · medium · numerical
For a caliper with 1 MSD = 0.5 mm and 20 VSD = 19 MSD, when jaws touch, the zero of the Vernier lies to the left of the main scale zero and the 14th VSD aligns. When measuring the length of a wire, MSR = 8.5 mm and the 10th VSD aligns. What is the true length of the wire?
A. 8.90 mm ✓ Correct
B. 8.40 mm
C. 8.60 mm
D. 8.75 mm
Solution: LC = 0.025 mm. Zero error = −(20 − 14) × 0.025 = −0.15 mm. Measured = 8.5 + (10 × 0.025) = 8.75 mm. True = 8.75 − (−0.15) = 8.90 mm.
Q11 — Measuring Instruments · medium · numerical
In an unusual Vernier caliper, 10 Vernier scale divisions coincide with 11 main scale divisions. If 1 MSD = 1 mm, what is the magnitude of the least count of this instrument?
A. 1.1 mm
B. 0.2 mm
C. 0.01 mm
D. 0.1 mm ✓ Correct
Solution: 10 VSD = 11 MSD → 1 VSD = 1.1 mm. LC = |1 MSD − 1 VSD| = |1.0 − 1.1| = 0.1 mm.
Q12 — Measuring Instruments · medium · numerical
The side of a metallic cube is measured with a Vernier caliper (1 MSD = 1 mm, 10 VSD = 9 MSD). When jaws close, the 2nd VSD coincides and the Vernier zero is to the right. The measured edge gives MSR = 1.0 cm with 5th VSD coincidence. The true volume of the cube is closest to:
A. 1.000 cm³
B. 1.030 cm³
C. 1.093 cm³ ✓ Correct
D. 1.158 cm³
Solution: Zero error = +(2 × 0.01) = +0.02 cm. Measured edge = 1.0 + (5 × 0.01) = 1.05 cm. True edge = 1.05 − 0.02 = 1.03 cm. Volume = (1.03)³ ≈ 1.093 cm³.
Q13 — Measuring Instruments · medium · numerical
A student measures the diameter of a circular disc with a Vernier caliper (1 MSD = 1 mm, 10 VSD = 9 MSD). Zero error: the Vernier zero is to the left of the main zero and the 8th VSD coincides when jaws touch. Measurement: MSR = 1.9 cm, 6th VSD coincides. What is the true area of cross-section of the disc? (Take π = 3.14)
A. 3.27 cm²
B. 3.08 cm² ✓ Correct
C. 2.95 cm²
D. 3.14 cm²
Solution: Zero error = −(10 − 8) × 0.01 = −0.02 cm. Measured d = 1.9 + (6 × 0.01) = 1.96 cm. True d = 1.96 − (−0.02) = 1.98 cm, so r = 0.99 cm. Area = πr² = 3.14 × (0.99)² ≈ 3.08 cm².
Q14 — Measuring Instruments · easy · theory
If ZE represents Zero Error and ZC represents Zero Correction, which relation is strictly correct?
A. ZC = 1/ZE
B. ZC = |ZE|
C. ZC = ZE
D. ZC = −ZE ✓ Correct
Solution: Zero correction is equal in magnitude and opposite in sign to the zero error: ZC = −ZE. True value = measured value + ZC = measured value − ZE.
Q15 — Measuring Instruments · medium · numerical
The depth of a beaker is measured using the depth gauge of a Vernier caliper (1 MSD = 1 mm, 10 VSD = 9 MSD). The zero error of the caliper is +0.02 cm. If MSR = 4.3 cm and the 7th VSD coincides, the correct depth of the beaker is:
A. 4.39 cm
B. 4.37 cm
C. 4.32 cm
D. 4.35 cm ✓ Correct
Solution: Measured depth = 4.3 + (7 × 0.01) = 4.37 cm. True depth = 4.37 − 0.02 = 4.35 cm.
Q16 — Measuring Instruments · medium · theory
If 1 MSD = a units and N divisions of the Vernier scale equal (N − m) divisions of the main scale, the least count of the Vernier caliper is:
A. $\frac{a}{m\,N}$
B. $\frac{a}{N}$
C. $\frac{m\,a}{N}$ ✓ Correct
D. $\frac{(N-m)\,a}{N}$
Solution: 1 VSD = ((N − m)/N)·a. LC = a − ((N − m)/N)·a = (m·a)/N.
Q17 — Measuring Instruments · medium · numerical
In a Vernier caliper, 1 MSD = 0.5 mm and the Vernier scale has 25 divisions coinciding with 24 main scale divisions. When jaws are closed, the Vernier zero is to the right of the main scale zero and the 5th division coincides. While measuring a bead, MSR = 0.5 cm and the 15th VSD coincides. The true diameter of the bead is:
A. 0.520 cm ✓ Correct
B. 0.500 cm
C. 0.530 cm
D. 0.510 cm
Solution: LC = 0.5 mm / 25 = 0.02 mm = 0.002 cm. Zero error = +(5 × 0.002) = +0.010 cm. Measured = 0.5 + (15 × 0.002) = 0.530 cm. True = 0.530 − 0.010 = 0.520 cm.
Q18 — Measuring Instruments · medium · theory
A cylinder is measured with a Vernier caliper (LC = 0.01 cm, zero error = −0.02 cm). Length: MSR = 4.9 cm, VSR = 4. Radius: MSR = 1.4 cm, VSR = 8. If the mass of the cylinder is 50.0 g, the true density of the material is closest to:
A. 1.52 g/cm³
B. 1.43 g/cm³ ✓ Correct
C. 1.61 g/cm³
D. 1.35 g/cm³
Solution: True length = (4.9 + 4×0.01) − (−0.02) = 4.94 + 0.02 = 4.96 cm. True radius = (1.4 + 8×0.01) + 0.02 = 1.50 cm. Volume = πR²L = 3.1416 × 1.50² × 4.96 ≈ 35.06 cm³. Density = 50.0 / 35.06 ≈ 1.43 g/cm³.
Q19 — Measuring Instruments · easy · theory
To increase the precision (decrease the least count) of a standard Vernier caliper without changing the main scale division length, one should:
A. Decrease the number of divisions on the Vernier scale
B. Increase the number of divisions on the Vernier scale ✓ Correct
C. Increase the thickness of the jaws
D. Decrease the main scale reading
Solution: LC = (1 MSD)/N. Increasing the number of Vernier divisions N decreases the least count, hence increasing precision.
Q20 — Measuring Instruments · medium · numerical
In a Vernier caliper, 10 VSD = 9 MSD and 1 MSD = 1 mm. When jaws touch, the Vernier zero lies between the 0 and 1 mm marks with the 4th division coinciding. When an object is held between the jaws, the Vernier zero lies between 3.4 cm and 3.5 cm and the 8th division coincides. The true length of the object is:
A. 3.42 cm
B. 3.48 cm
C. 3.52 cm
D. 3.44 cm ✓ Correct
Solution: Zero error = +(4 × 0.01) = +0.04 cm. Measured = 3.4 + (8 × 0.01) = 3.48 cm. True = 3.48 − 0.04 = 3.44 cm.