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  4. 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 :

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Q. 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 :

JEE MainJEE Main 2018Physical World, Units and Measurements

Solution:

Given: Linear distance covered in 5 rotations $=0.25\, cm$
So, distance $=\frac{0.25}{5}=0.05\, cm$
There are 100 circular scale divisions.
Therefore, least count $=0.05 \times 10^{-2}\, cm .$
Reading from 4 main scale divisions
$\Rightarrow 4 \times 0.05=0.2\, cm$
Reading from 30 circular scale divisions
$\Rightarrow 30 \times 0.05 \times 10^{-2}=1.5 \times 10^{-2}\, cm$
Thus, thickness of wire $=0.2\, cm +1.5 \times 10^{-2}=0.2150\, cm$