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Screw gauge: thickness/diameter of sheet or wire: NEET MCQs with answers

67 questions on this topic of Experimental Skills (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. In an experiment to measure the diameter of a uniform metallic wire using a screw gauge, a student takes measurements along two mutually perpendicular directions at the same cross-section. This specific practice is recommended primarily to:

  1. eliminate the zero error of the screw gauge.
  2. eliminate backlash error between the screw and the nut.
  3. account for any non-uniformity or non-circular cross-section of the wire.
  4. avoid deformation of the wire due to excessive pressure applied by the spindle.
Show answer and explanation

Answer: (C) account for any non-uniformity or non-circular cross-section of the wire.

Wires are often slightly elliptical or irregular rather than perfectly cylindrical. Measuring diameter along two perpendicular directions at the same cross-section ensures that any ovality or non-circular cross-section is averaged out, giving a more accurate mean diameter.

medium

2. A screw gauge has a pitch of 0.5 mm and 50 circular-scale divisions. The main-scale reading is 4 mm and the 10th circular division coincides with the reference line (no zero error). The diameter of the wire is:

  1. 4.2 mm
  2. 5 mm
  3. 4.1 mm
  4. 4.5 mm
Show answer and explanation

Answer: (C) 4.1 mm

Least count = pitch/divisions = 0.5/50 = 0.01 mm. Diameter = 4 + 10 × 0.01 = 4.1 mm.

hard

3. A student measures the diameter of a thin wire using a screw gauge. The pitch is 0.5 mm and the head scale has 100 divisions. The initial zero reading shows the 95th division on the circular scale aligning with the reference line (negative error). When measuring the wire, the main scale reads 1.0 mm and the 45th division coincides with the reference line. What is the measured cross-sectional area of the wire? (Take π = 3.142)

  1. 1.77 mm²
  2. 0.95 mm²
  3. 1.23 mm²
  4. 2.14 mm²
Show answer and explanation

Answer: (C) 1.23 mm²

Least count = 0.5 / 100 = 0.005 mm. Zero error = -(100 - 95) × 0.005 mm = -0.025 mm. Measured diameter = 1.0 mm + 45 × 0.005 mm - (-0.025 mm) = 1.0 + 0.225 + 0.025 = 1.250 mm. Cross-sectional area = π(d/2)² = 3.142 × (1.25 / 2)² = 3.142 × 0.3906 = 1.227 mm² ≈ 1.23 mm².

medium

4. A screw gauge has a pitch of 1 mm and 100 circular-scale divisions. The main-scale reading is 6 mm and the 33rd circular division coincides with the reference line (no zero error). The diameter of the wire is:

  1. 6.33 mm
  2. 9.3 mm
  3. 12.66 mm
  4. 0.33 mm
Show answer and explanation

Answer: (A) 6.33 mm

Least count = pitch/divisions = 1/100 = 0.01 mm. Diameter = 6 + 33 × 0.01 = 6.33 mm.

easy

5. Which of the following procedures helps in minimizing backlash error while taking measurements using a screw gauge?

  1. Cleaning the screw threads with lubricating oil before every observation
  2. Taking readings only along the vertical diameter of a wire
  3. Decreasing the total number of circular scale divisions
  4. Always rotating the screw head in the same direction when advancing towards the object
Show answer and explanation

Answer: (D) Always rotating the screw head in the same direction when advancing towards the object

Backlash error arises due to wear and tear or loose fitting between the screw and the internal nut threads. It is minimized by consistently turning the thimble in one direction only while making a setting.

hard

6. The diameter of a wire measured with a screw gauge is recorded as 1.25 mm. If the pitch of the screw gauge is 0.5 mm and it contains 50 divisions on its circular scale, the maximum possible fractional error in the cross-sectional area of the wire is:

  1. 0.008
  2. 0.016
  3. 0.024
  4. 0.032
Show answer and explanation

Answer: (B) 0.016

Least count Δd = 0.5 mm / 50 = 0.01 mm. The cross-sectional area is A = πd²/4. Therefore, ΔA/A = 2(Δd/d) = 2 × (0.01 mm / 1.25 mm) = 2 × 0.008 = 0.016.

hard

7. A student measures the diameter of a wire as d = 1.25 mm using a screw gauge with least count 0.01 mm, and its length as L = 10.0 cm using a meter scale with least count 0.1 cm. The percentage error in the calculated volume of the wire is:

  1. 1.6%
  2. 2.6%
  3. 1.8%
  4. 3.2%
Show answer and explanation

Answer: (B) 2.6%

Volume of wire V = π(d/2)²L = (π/4) d² L. The fractional error is ΔV/V = 2(Δd/d) + (ΔL/L). Here, Δd = 0.01 mm, d = 1.25 mm, ΔL = 0.1 cm, L = 10.0 cm. Thus, ΔV/V = 2(0.01 / 1.25) + (0.1 / 10.0) = 2(0.008) + 0.010 = 0.016 + 0.010 = 0.026. Therefore, the percentage error is 0.026 × 100% = 2.6%.

medium

8. A screw gauge has a pitch of 1 mm and 50 circular-scale divisions. The main-scale reading is 8 mm and the 11th circular division coincides with the reference line (no zero error). The diameter of the wire is:

  1. 8.22 mm
  2. 9.1 mm
  3. 0.22 mm
  4. 10.2 mm
Show answer and explanation

Answer: (A) 8.22 mm

Least count = pitch/divisions = 1/50 = 0.02 mm. Diameter = 8 + 11 × 0.02 = 8.22 mm.

medium

9. In a specific screw gauge, the pitch is 0.5 mm and there are 50 divisions on the circular scale. While measuring the diameter of a thin copper wire, the main scale reads 0.5 mm and the circular scale reads 20 divisions. If the zero error of this instrument is -0.04 mm, what is the cross-sectional area of the wire? (Use π = 3.14)

  1. 0.48 mm²
  2. 0.32 mm²
  3. 0.55 mm²
  4. 0.43 mm²
Show answer and explanation

Answer: (D) 0.43 mm²

Least count = 0.5 mm / 50 = 0.01 mm. Measured reading = 0.5 mm + (20 × 0.01 mm) = 0.70 mm. True diameter = Measured reading - Zero error = 0.70 mm - (-0.04 mm) = 0.74 mm. Cross-sectional area A = π × (d/2)² = 3.14 × (0.37 mm)² = 3.14 × 0.1369 mm² ≈ 0.43 mm².

medium

10. A screw gauge has a least count of 0.01 mm and a zero error of -0.04 mm. While measuring the thickness of a metallic foil, the main scale reads 2.0 mm and the circular scale reads 28 divisions. The true thickness of the foil is:

  1. 2.24 mm
  2. 2.28 mm
  3. 2.32 mm
  4. 2.36 mm
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Answer: (C) 2.32 mm

Observed reading = 2.0 mm + (28 × 0.01 mm) = 2.28 mm. True reading = Observed reading - Zero error = 2.28 mm - (-0.04 mm) = 2.32 mm.

medium

11. The thickness of a uniform aluminum sheet is measured using a screw gauge. The pitch is 0.5 mm and the head scale contains 50 divisions. When the sheet is inserted, the main scale reading is 1.0 mm and the 38th head scale division coincides with the reference line. Before the experiment, when the studs were closed, the 47th head scale division coincided with the reference line. What is the corrected thickness of the sheet?

  1. 1.35 mm
  2. 1.41 mm
  3. 1.44 mm
  4. 1.38 mm
Show answer and explanation

Answer: (B) 1.41 mm

LC = 0.5 / 50 = 0.01 mm. Zero error: 47th division coincides, which means the zero is above the reference line by 3 divisions (50 - 47 = 3). Zero error = -3 × 0.01 mm = -0.03 mm. Observed reading = 1.0 mm + 38 × 0.01 mm = 1.38 mm. Corrected thickness = 1.38 - (-0.03) = 1.41 mm.

medium

12. A screw gauge has a pitch of 1 mm and 50 circular-scale divisions. The main-scale reading is 12 mm and the 7th circular division coincides with the reference line (no zero error). The diameter of the wire is:

  1. 13.4 mm
  2. 12.7 mm
  3. 12.14 mm
  4. 0.14 mm
Show answer and explanation

Answer: (C) 12.14 mm

Least count = pitch/divisions = 1/50 = 0.02 mm. Diameter = 12 + 7 × 0.02 = 12.14 mm.

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