JEE Advanced 2022 Paper 2 Q17 Physics Errors & Measurements Vernier & Screw Gauge Medium

JEE Advanced 2022 Paper 2 · Q17 · Vernier & Screw Gauge

Area of the cross-section of a wire is measured using a screw gauge. The pitch of the main scale is $0.5\,\text{mm}$. The circular scale has $100$ divisions and for one full rotation of the circular scale, the main scale shifts by two divisions. The measured readings are listed below.

| Measurement condition | Main scale reading | Circular scale reading | |---|---|---| | Two arms of gauge touching each other without wire | $0$ division | $4$ divisions | | Attempt-1: With wire | $4$ divisions | $20$ divisions | | Attempt-2: With wire | $4$ divisions | $16$ divisions |

What are the diameter and cross-sectional area of the wire measured using the screw gauge?

  1. A. $2.22 \pm 0.02\,\text{mm},\ \pi(1.23 \pm 0.02)\,\text{mm}^2$
  2. B. $2.22 \pm 0.01\,\text{mm},\ \pi(1.23 \pm 0.01)\,\text{mm}^2$
  3. C. $2.14 \pm 0.02\,\text{mm},\ \pi(1.14 \pm 0.02)\,\text{mm}^2$
  4. D. $2.14 \pm 0.01\,\text{mm},\ \pi(1.14 \pm 0.01)\,\text{mm}^2$
Reveal answer + step-by-step solution

Correct answer:C

Solution

Pitch of main scale $= 0.5\,\text{mm}$. One full rotation of circular scale advances the main scale by two divisions, i.e. by $2\times 0.5 = 1\,\text{mm}$. So effective pitch (linear travel per rotation) $= 1\,\text{mm}$; least count $= 1/100 = 0.01\,\text{mm}$. Zero error $= +4 \times 0.01 = +0.04\,\text{mm}$. Attempt-1 reading $= 4\times 0.5 + 20\times 0.01 = 2.00 + 0.20 = 2.20\,\text{mm}$; corrected $= 2.20 - 0.04 = 2.16\,\text{mm}$. Attempt-2 reading $= 4\times 0.5 + 16\times 0.01 = 2.00 + 0.16 = 2.16\,\text{mm}$; corrected $= 2.16 - 0.04 = 2.12\,\text{mm}$. Mean diameter $d = (2.16 + 2.12)/2 = 2.14\,\text{mm}$. Mean absolute deviation $\Delta d = (0.02 + 0.02)/2 = 0.02\,\text{mm}$. Area $A = \pi d^2/4 = \pi (2.14)^2/4 \approx \pi \cdot 1.14\,\text{mm}^2$; $\Delta A/A = 2\Delta d/d = 2\cdot 0.02/2.14 \approx 0.019$, so $\Delta A \approx 0.02\pi\,\text{mm}^2$. Result: $d = 2.14 \pm 0.02\,\text{mm}$, $A = \pi(1.14 \pm 0.02)\,\text{mm}^2$ — option (C).

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