Area of the cross-section of a wire is measur | vedclass

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(C) Pitch $= 0.5 	ext{ mm}$. Since $1$ rotation shifts the main scale by $2$ divisions, the pitch is $2 	imes 0.5 	ext{ mm} = 1.0 	ext{ mm}$.Least

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Pitch $= 0.5 	ext{ mm}$. Since $1$ rotation shifts the main scale by $2$ divisions, the pitch is $2 	imes 0.5 	ext{ mm} = 1.0 	ext{ mm}$.Least Count $(LC)$ $= \frac{	ext{Pitch}}{	ext{Total divisions}} = \frac{1.0 	ext{ mm}}{100} = 0.01 	ext{ mm}$.Zero Error $= +4 	imes 0.01 	ext{ mm} = +0.04 	ext{ mm}$.Reading $1 = (4 	imes 0.5 	ext{ mm}) + (20 	imes 0.01 	ext{ mm}) - 0.04 	ext{ mm} = 2.0 + 0.20 - 0.04 = 2.16 	ext{ mm}$.Reading $2 = (4 	imes 0.5 	ext{ mm}) + (16 	imes 0.01 	ext{ mm}) - 0.04 	ext{ mm} = 2.0 + 0.16 - 0.04 = 2.12 	ext{ mm}$.Mean Diameter $(d) = \frac{2.16 + 2.12}{2} = 2.14 	ext{ mm}$.Mean absolute error in $d = \frac{|2.16 - 2.14| + |2.12 - 2.14|}{2} = 0.02 	ext{ mm}$.Diameter $= 2.14 \pm 0.02 	ext{ mm}$.Area $A = \frac{\pi d^2}{4} = \frac{\pi (2.14)^2}{4} = \pi (1.1449) \approx 1.14 \pi 	ext{ mm}^2$.Relative error in $A = 2 	imes \frac{\Delta d}{d} = 2 	imes \frac{0.02}{2.14} \approx 0.0187$.Absolute error in $A = 0.0187 	imes 1.14 \approx 0.02 	ext{ mm}^2$.Thus, Area $= \pi(1.14 \pm 0.02) 	ext{ mm}^2$.

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

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