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JEE Main 2026 (23 Jan, Shift 1) · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2026 (23 Jan, Shift 1)

Q8JEE Main 2026 · 23 JAN · SHIFT 1Medium

Least Count, Vernier Callipers & Screw Gauge

In a screw gauge, the zero of the circular scale lies 3 divisions above the horizontal pitch line when their metallic studs are brought in contact. Using this instrument thickness of a sheet is measured. If pitch scale reading is 1 mm and the circular scale reading is 51 then the correct thickness of the sheet is ______ mm.

[Assume least count is 0.01 mm ]

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Answer: (a)

  1. Zero of circular scale above the pitch line means negative zero error =−3×0.01=−0.03 mm= -3\times 0.01 = -0.03\text{ mm}
  2. Observed reading =1+51×0.01=1.51 mm= 1 + 51\times 0.01 = 1.51\text{ mm}
  3. Correct thickness =1.51−(−0.03)=1.54 mm= 1.51 - (-0.03) = 1.54\text{ mm}. So the answer is (a).
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