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 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. $2.22 \pm 0.02 mm , \pi(1.23 \pm 0.02) mm ^2$
• B. $2.22 \pm 0.01 mm , \pi(1.23 \pm 0.01) mm ^2$
• C. $2.14 \pm 0.02 mm , \pi(1.14 \pm 0.02) mm ^2$
• D. $2.14 \pm 0.01 mm , \pi(1.14 \pm 0.01) mm ^2$

Answer 1

Answer: (C)

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

Answer 2

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 =$\frac{0.02+0.02}{2}=0.02$
Diameter =$2.14±0.02$
Area =$4\pi d^2$
Therefore correct answer is $2.14 \pm 0.02 mm , \pi(1.14 \pm 0.02) mm ^2$