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 division 20division Attempt-2: With wire 4 division 16 division What are the diameter and cross-sectional area of the wire measured using the screw gauge?

Question

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 division 20division Attempt-2: With wire 4 division 16 division What are the diameter and cross-sectional area of the wire measured using the screw gauge? (A) 2.22 ± 0.02 mm , π(1.23 ± 0.02) mm 2 (B) 2.22 ± 0.01 mm , π(1.23 ± 0.01) mm 2 (C) 2.14 ± 0.02 mm , π(1.14 ± 0.02) mm 2 (D) 2.14 ± 0.01 mm , π(1.14 ± 0.01) mm 2

Tardigrade Admin · Accepted Answer

LC =(0.1/100)=0.001 mm Zero error =4 × 0.001=0.004 mm Reading 1=0.5 × 4+20 × 0.001-0.004=2.16 mm Reading 2=0.5 × 4+16 × 0.001-0.004=2.12 mm Mean value =2.14 mm Mean absolute error =(0.02+0.02/2)=0.02 Diameter =2.14 ± 0.02 Area =(π/4) d 2

$2.22 \pm 0.02 mm, π (1.23 \pm 0.02) m m^{2}$

$2.22 \pm 0.01 mm, π (1.23 \pm 0.01) m m^{2}$

$2.14 \pm 0.02 mm, π (1.14 \pm 0.02) m m^{2}$

$2.14 \pm 0.01 mm, π (1.14 \pm 0.01) m m^{2}$

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 division	20division
Attempt-2: With wire	4 division	16 division

2.22±0.02mm,π(1.23±0.02)mm2

2.22±0.01mm,π(1.23±0.01)mm2

2.14±0.02mm,π(1.14±0.02)mm2

2.14±0.01mm,π(1.14±0.01)mm2

LC=1000.1​=0.001mm Zero error =4×0.001=0.004mm Reading 1=0.5×4+20×0.001−0.004=2.16mm Reading 2=0.5×4+16×0.001−0.004=2.12mm Mean value =2.14mm Mean absolute error =20.02+0.02​=0.02 Diameter =2.14±0.02 Area =4π​d2

$2.22 \pm 0.02 mm, π (1.23 \pm 0.02) m m^{2}$

$2.22 \pm 0.01 mm, π (1.23 \pm 0.01) m m^{2}$

$2.14 \pm 0.02 mm, π (1.14 \pm 0.02) m m^{2}$

$2.14 \pm 0.01 mm, π (1.14 \pm 0.01) m m^{2}$