A screw gauge has 50 divisions on its circular scale. The pitch of the screw gauge is 0.5 mm. When the jaws are closed, the 4th division of the circular scale is above the reference line and the zero of the main scale is clearly visible. When a wire is placed between the jaws, the main scale reading is 4 mm and the 24th division of the circular scale coincides with the reference line. The correct diameter of the wire is:

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Appears in NEET UG
Solution ✔ Verified
  • A4.20 mm
  • B4.22 mm
  • C4.24 mm
  • D4.26 mm
Explanation

The least count (LC) = Pitch / Number of divisions on circular scale = 0.5 mm / 50 = 0.01 mm. Zero error (ZE) = + (4 divisions) × LC = + 4 × 0.01 mm = + 0.04 mm. Observed reading (OR) = Main Scale Reading + Circular Scale Reading × LC = 4 mm + 24 × 0.01 mm = 4 mm + 0.24 mm = 4.24 mm. Correct Reading = OR - ZE = 4.24 mm - (+0.04 mm) = 4.20 mm. Let me recheck this for calculation. ZE = 4 * 0.01 = 0.04. OR = 4 + 24*0.01 = 4.24. Correct = 4.24 - 0.04 = 4.20. So A=4.20mm. My current answer is B=4.22mm. This is a mismatch. I will correct the answer to A.A screw gauge has 50 divisions on its circular scale. The pitch of the screw gauge is 0.5 mm. When the jaws are closed, the 4th division of the circular scale is above the reference line and the zero of the main scale is clearly visible. When a wire is placed between the jaws, the main scale reading is 4 mm and the 24th division of the circular scale coincides with the reference line. The correct diameter of the wire is: A: 4.20 mm B: 4.22 mm C: 4.24 mm D: 4.26 mm Answer: A Explanation:

The least count (LC) = Pitch / Number of divisions on circular scale = 0.5 mm / 50 = 0.01 mm. The zero error (ZE) = + (4 divisions) × LC = + 4 × 0.01 mm = + 0.04 mm. The observed reading (OR) = Main Scale Reading + Circular Scale Reading × LC = 4 mm + 24 × 0.01 mm = 4 mm + 0.24 mm = 4.24 mm. The correct reading = OR - ZE = 4.24 mm - (+0.04 mm) = 4.20 mm.

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