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

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?

TestHub · Accepted Answer

Note: This question is given bonus by JEE council.
In one rotation 2 divisions of the main scale are crossed. Therefore, the least count of the screw gauge is
$L C = \frac{2 \times 0.5}{100} = 0.01 mm$
And the zero error is $4 \times L C = 0.04 mm$
Reading-1
$R_{1} = MSR + LC \times CSR - (Zero error)$
$= (2 + 0.20 - 0.04) mm$
$= 2.16 mm$
Reading-2
$R_{2} = (2 + 0.16 - 0.04) mm = 2.12 mm$
Therefore, average reading $R_{m} = \frac{R_{1} + R_{2}}{2} = 2.14 mm$
Average mean error $= \frac{|R_{m} - R_{1}| + |R_{m} - R_{2}|}{2} = 0.02 mm$
$\Rightarrow$ Diameter $d = (2.14 \pm 0.02) mm$
Area $= \frac{π d^{2}}{4}$
$\Rightarrow ∆ Area = \frac{2 π d}{4} (∆ d)$
$\Rightarrow ∆ Area ≅ π (0.02)$
$\Rightarrow Area = \frac{π d^{2}}{4} \pm ∆ Area = π (1.14 \pm 0.02) {mm}^{2}$

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