JEE Advanced 2021 Paper 1 Q01 Physics Errors & Measurements Vernier & Screw Gauge Medium

JEE Advanced 2021 Paper 1 · Q01 · Vernier & Screw Gauge

The smallest division on the main scale of a Vernier calipers is $0.1$ cm. Ten divisions of the Vernier scale correspond to nine divisions of the main scale, giving a least count of $0.01$ cm. When the two jaws are in contact (zero-error check), the Vernier zero lies slightly ahead of the main-scale zero, with the 6th Vernier division aligning with a main-scale division — giving a negative zero error of $-0.04$ cm. When a solid sphere is held between the jaws (measurement reading), the main-scale reading is $3.1$ cm and the 1st Vernier division aligns — giving an observed reading of $3.11$ cm. The correct diameter of the sphere is

  1. A. 3.07 cm
  2. B. 3.11 cm
  3. C. 3.15 cm
  4. D. 3.17 cm
Reveal answer + step-by-step solution

Correct answer:C

Solution

Least count = (1 − 9/10) × 0.1 = 0.01 cm. From the left figure (no gap), the Vernier zero lies past the main-scale zero with the 6th Vernier division aligning, so the zero error is positive: ZE = 0 + 6 × 0.01 − 0.1 (vernier reads at −0.1 main-scale position) actually evaluates to −0.04 cm (negative zero error). From the right figure, observed reading = 3.1 + 1 × 0.01 = 3.11 cm. Corrected diameter = observed − zero error = 3.11 − (−0.04) = 3.15 cm.

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