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 division | | 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?

(C) $2.14 \pm 0.02 mm, \pi(1.14 \pm 0.02) mm^{2}$

(A) $2.22 \pm 0.02 mm, \pi(1.23 \pm 0.02) mm^{2}$

(B) $2.22 \pm 0.01 mm, \pi(1.23 \pm 0.01) mm^{2}$

(D) $2.14 \pm 0.01 mm, \pi(1.14 \pm 0.01) mm^{2}$

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Units & Measurement

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

Answer: (C)

Solution

Least count:

L C = \frac{pitch}{divisions} = \frac{0.5 \times 2}{100} = 0.01

mm

Zero error:

4 \times 0.01 = 0.04

mm

Reading 1:

4 \times 0.5 + 20 \times 0.01 - 0.04 = 2.16

mm

Reading 2:

4 \times 0.5 + 16 \times 0.01 - 0.04 = 2.12

mm

Mean:

\frac{2.16 + 2.12}{2} = 2.14

mm

Mean absolute error:

\frac{∣2.16 - 2.14∣ + ∣2.12 - 2.14∣}{2} = 0.02

mm

Diameter:

2.14 \pm 0.02

mm

Area:

A = \frac{π d ^{2}}{4} = π (\frac{2.14}{2})^{2} = π (1.14)^{2}

mm²

With error propagation:

\frac{Δ A}{A} = 2 \frac{Δ d}{d}

, giving area as

π (1.14 \pm 0.02)

mm²

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Single correct option

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JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Units & Measurement

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