JEE Advanced 2022 — Paper 2 — PHY-III solution

JEE Advanced2022Paper 2PHY-III

Area of the cross-section of a wire is measured using a screw gauge. The pitch of the main scale is 0.5 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?

A
$2.22 \pm 0.02$ mm, $π (1.23 \pm 0.02)$ mm $^{2}$
B
$2.22 \pm 0.01$ mm, $π (1.23 \pm 0.01)$ mm $^{2}$
C
$2.14 \pm 0.02$ mm, $π (1.14 \pm 0.02)$ mm $^{2}$
D
$2.14 \pm 0.01$ mm, $π (1.14 \pm 0.01)$ mm $^{2}$

Solution

One full rotation moves the main scale by 2 divisions $\times 0.5$ mm = 1 mm. With 100 circular divisions, the least count is $1/100 = 0.01$ mm.

Zero error: with no wire, MSR = 0, CSR = 4 divisions, so zero error $= + 4 \times 0.01 = 0.04$ mm (positive, must be subtracted).

Attempt-1: MSR $= 4 \times 0.5 = 2$ mm; CSR $= 20 \times 0.01 = 0.20$ mm. Corrected: $R_{1} = 2 + 0.20 - 0.04 = 2.16$ mm.

Attempt-2: MSR $= 2$ mm; CSR $= 16 \times 0.01 = 0.16$ mm. Corrected: $R_{2} = 2 + 0.16 - 0.04 = 2.12$ mm.

Mean diameter: $d = (R_{1} + R_{2}) /2 = 2.14$ mm. Mean absolute error: $\frac{∣ R _{1} - d ∣ + ∣ R _{2} - d ∣}{2} = \frac{0.02 + 0.02}{2} = 0.02$ mm.

Diameter $= (2.14 \pm 0.02)$ mm. Area $A = π d^{2} /4 = π (2.14)^{2} /4 \approx π \cdot 1.14$ mm $^{2}$ . Uncertainty: $Δ A = \frac{π \cdot 2 d \cdot Δ d}{4} = π \cdot \frac{2.14 \cdot 0.02}{2} \approx π \cdot 0.02$ mm $^{2}$ . Hence area $= π (1.14 \pm 0.02)$ mm $^{2}$ . Option (C).

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