Question 264: Units , Dimensions and Error Analysis

Question

Area of the cross-section of a wire is measured using a screw gauge. The pitch of the main scale is $0.5 \mathrm{~mm}$. 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?

Accepted Answer

$$
\begin{aligned}
& 	ext {LC }=\frac{0.1}{100}=0.001 \mathrm{~mm} \\
& 	ext {Zero error }=4 	imes 0.001=0.004 \mathrm{~mm} \\
& 	ext {Reading } 1=0.5 	imes 4+20 	imes 0.001-0.004=2.16 \mathrm{~mm} \\
& 	ext {Reading } 2=0.5 	imes 4+16 	imes 0.001-0.004=2.12 \mathrm{~mm} \\
& 	ext {Mean value }=2.14 \mathrm{~mm} \\
& 	ext {Mean absolute error }=\frac{0.02+0.02}{2}=0.02 \\
& 	ext {Diameter }=2.14 \pm 0.02 \\
& 	ext {Area }=\frac{\pi}{4} \mathrm{~d}^2
\end{aligned}
$$

Answer

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

Answer

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

Answer

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

Answer

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

Units , Dimensions and Error Analysis

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