In a screw gauge, 5 complete rotations of screw cause it to move a linear distance of 0.25, cm . The

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1.Units, Dimensions and Measurement

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In a screw gauge, $5$ complete rotations of the screw cause it to move a linear distance of $0.25\, cm$. There are $100$ circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of $4$ main scale divisions and $30$ circular scale divisions . Assuming negligible zero error, the thickness of the wire is

A

$0.0430\,cm$

B

$0.3150\,cm$

C

$0.43 00\,cm$

D

$0.2150\, cm$

(JEE MAIN-2018)

Solution

In one rotation scale moves $\frac{0.25}{5}=0.05 cm$

Least count $=0.05 \times 10^{-2} cm$

For 4 main scale division $=4 \times 0.05=0.2 cm$

For circular scale divosion $=30 \times 0.05 \times 10^{-2}=1.5 \times 10^{-2} cm$

Thickness of wire $=0.2+0.015=0.2150 cm$

Standard 11

Physics

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