Question

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. $0.4300\, cm$
• B. $0.2150\, cm$
• C. $0.3150\, cm$
• D. $0.0430\, cm$

Answer 1

Answer: (B)

$0.2150\, cm$

Answer 2

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 $=0.2\, cm +1.5 \times 10^{-2}=0.2150\, cm$