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In 10 complete rotations, the distance moved on the linear scale is $1 cm$ . There are 100 divisions on circular scale. When the gap AB is just closed, the 95 th division of circular scale coincides with the reference line. While measuring the diameter of a wire, the linear scale reads $2 mm$ and 45 th division on the circular scale coincides with the reference line. Find cross-sectional area of the wire.

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$0.45 {cm}^{2}$

$3.8 {cm}^{2}$

$0.049 {cm}^{2}$

$0.098 {cm}^{2}$

answer is B.

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Detailed Solution

Here, Pitch = $\frac{1 cm}{10} = \frac{10 mm}{10} = 1 mm$

$(L C) = \frac{Pitch}{Number of divisions on circular scale}$

$= \frac{1}{100} = 0.01 mm$

Zero error = $- (100 - 95) \times LC$

$= - 5 \times 0.01 = - 0.05 mm$

The diameter of wire is

d= Reading of main scale + Reading of circular scale + (Zero error)

$= 2 mm + (45 \times 0.01 + 0.05) mm$

$= 2.5 mm$

Cross-sectional area = $\frac{π d^{2}}{4}$

$= \frac{3.14 \times {(2.5 \times 10^{- 1})}^{2}}{4} {cm}^{2}$

$= 0.049 {cm}^{2}$

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