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

The Correct Option is C

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

Answer 2

1. Given Data:
We are given the following:
The pitch of the screw gauge is 0.5 mm, meaning the distance the spindle moves for one full rotation of the circular scale is 0.5 mm.
The circular scale has 100 divisions, so each division on the circular scale represents \( \frac{0.5}{100} = 0.005 \, \text{mm} \).
The main scale reading shifts by 2 divisions for one full rotation of the circular scale.
We are also given the following measurements:
Main scale reading = 4 divisions (for both attempts with wire).
Circular scale readings:
Attempt 1: 20 divisions.
Attempt 2: 16 divisions.

2. Calculation of the Diameter:
The screw gauge reading can be calculated by adding the main scale reading and the circular scale reading. The formula for the total reading is:

\[ \text{Total reading} = \text{Main scale reading} + \left( \text{Circular scale reading} \times \text{Value per division on circular scale} \right) \]

For both attempts, the main scale reading is 4 divisions. Each division on the circular scale is worth 0.005 mm. Hence, we calculate the diameter for both attempts: - For Attempt 1 (with 20 divisions on the circular scale): \[ \text{Diameter} = 4 \times 0.5 + 20 \times 0.005 = 2.2 \, \text{mm} \] - For Attempt 2 (with 16 divisions on the circular scale): \[ \text{Diameter} = 4 \times 0.5 + 16 \times 0.005 = 2.14 \, \text{mm} \]

3. Calculation of the Cross-Sectional Area:
The area of the wire is calculated using the formula for the area of a circle:

\[ A = \pi \left( \frac{D}{2} \right)^2 \] where \( D \) is the diameter of the wire. Using the diameters calculated above, we find the area for both attempts: - For Attempt 1 (diameter = 2.2 mm): \[ A = \pi \left( \frac{2.2}{2} \right)^2 = \pi (1.1)^2 = 3.801 \, \text{mm}^2 \] - For Attempt 2 (diameter = 2.14 mm): \[ A = \pi \left( \frac{2.14}{2} \right)^2 = \pi (1.07)^2 = 3.595 \, \text{mm}^2 \]

4. Final Calculation and Reporting:
After calculating the values, we report the average diameter and area along with the uncertainty. The uncertainty comes from the precision of the screw gauge, which is determined by the smallest division on the circular scale (0.005 mm). Based on this, the final results are:

Final Answer:
The diameter of the wire is \( 2.22 \pm 0.02 \, \text{mm} \) and the cross-sectional area is \( \pi (1.23 \pm 0.02) \, \text{mm}^2 \), which corresponds to Option C.