Verified Solution • 2025–26 Session moderate

QJEEAdv22Ph7db1: JEE Advanced 2022 Physics - Class 11 JEEAdv22Ph7db1 Solution: Area of the cross-section of a wire is measured using a screw…

Area of the cross-section of a wire is measured using a screw gauge. The pitch of the main scale is 0.5 mm0.5 \mathrm{~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?

Multiple Choice Options

  • A 2.22±0.02 mm,π(1.23±0.02)mm22.22 \pm 0.02 \mathrm{~mm}, \pi(1.23 \pm 0.02) \mathrm{mm}^{2}
  • B 2.22±0.01 mm,π(1.23±0.01)mm22.22 \pm 0.01 \mathrm{~mm}, \pi(1.23 \pm 0.01) \mathrm{mm}^{2}
  • C 2.14±0.02 mm2.14 \pm 0.02 \mathrm{~mm} , π(1.14±0.02)mm2\pi(1.14 \pm 0.02) \mathrm{mm}^{2}
  • D 2.14±0.01 mm,π(1.14±0.01)mm22.14 \pm 0.01 \mathrm{~mm}, \pi(1.14 \pm 0.01) \mathrm{mm}^{2}

Step-by-Step Verified Solution

amp;LC =0.1100=0.001 mmamp;Zero error =4×0.001=0.004 mmamp;Reading 1=0.5×4+20×0.001−0.004=2.16 mmamp;Reading 2=0.5×4+16×0.001−0.004=2.12 mmamp;Mean value =2.14 mmamp;Mean absolute error =0.02+0.022=0.02amp;Diameter =2.14±0.02amp;Area =π4 d2\begin{aligned}& \text {LC }=\frac{0.1}{100}=0.001 \mathrm{~mm} \\\\& \text {Zero error }=4 \times 0.001=0.004 \mathrm{~mm} \\\\& \text {Reading } 1=0.5 \times 4+20 \times 0.001-0.004=2.16 \mathrm{~mm} \\\\& \text {Reading } 2=0.5 \times 4+16 \times 0.001-0.004=2.12 \mathrm{~mm} \\\\& \text {Mean value }=2.14 \mathrm{~mm} \\\\& \text {Mean absolute error }=\frac{0.02+0.02}{2}=0.02 \\\\& \text {Diameter }=2.14 \pm 0.02 \\\\& \text {Area }=\frac{\pi}{4} \mathrm{~d}^2\end{aligned}

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