Area of the cross-section of a wire is measured using a screw...

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

$LC = \dfrac{2 \times 0 . 5}{100} = 0 . 01mm$

And the zero error is $4 \times LC = 0 . 04mm$

Reading-1

${R}_{1} = MSR + LC \times CSR - \left(Zeroerror\right)$

$= \left(2 + 0 . 20 - 0 . 04\right)mm$

$= 2 . 16mm$

Reading-2

${R}_{2} = \left(2 + 0 . 16 - 0 . 04\right)mm = 2 . 12mm$

Therefore, average reading ${R}_{m} = \dfrac{{R}_{1} + {R}_{2}}{2} = 2 . 14mm$

Average mean error $= \dfrac{\left|{R}_{m} - {R}_{1}\right| + \left|{R}_{m} - {R}_{2}\right|}{2} = 0 . 02mm$

$⇒$ Diameter $d = \left(2 . 14 \pm 0 . 02\right)mm$

Area $= \dfrac{\pi{d}^{2}}{4}$

$⇒ ∆ Area = \dfrac{2\pid}{4}\left(∆ d\right)$

$⇒ ∆ Area ≅ \pi\left(0 . 02\right)$

$⇒ Area = \dfrac{\pi{d}^{2}}{4} \pm ∆ Area = \pi\left(1 . 14 \pm 0 . 02\right){mm}^{2}$