#### A wire of unknown diameter is measured using a screw gauge. The screw gauge has $100$ divisions on its circular scale, and the pitch of the screw is $0.5\ mm$. The zero error of the screw gauge is $+0.02\ mm$. The wire is measured by placing it between the anvil and the spindle, and the thimble scale reading is $32.5$ divisions, while the circular scale reading is $48$ divisions. Calculate the diameter of the wire.Option: 1 $8.048\ mm$Option: 2 $80.48\ mm$Option: 3 $80.48\ cm$Option: 4 $804.8\ mm$

1. Determine the Total Reading of the Screw Gauge:
The total reading of the screw gauge is the sum of the thimble scale reading
and the circular scale reading. Given that the thimble scale reading is $32.5$
divisions and the circular scale reading is $48$ divisions, calculate the total

Total Reading $=$ Thimble Scale Reading $+$ Circular Scale Reading
Total Reading $=$$32.5$ divisions $+$ $48$ divisions
Total Reading $=80.5$ divisions

The actual reading of the screw gauge is the total reading corrected for the
zero error. Given that the zero error is $+0.02\ mm$ (positive because it’s
away from the zero mark), subtract the zero error from the total reading:

Actual Reading $=$Total Reading - Zero Error
Actual Reading $=80.5$ divisions $-0.02\ mm$
Actual Reading $=80.48$ divisions
3. Calculate the Distance Moved by the Screw:
The distance moved by the screw is the product of the actual reading and
the pitch of the screw. The pitch of the screw is $0.05\ mm$.
Distance Moved $=$Actual Reading $\times$Pitch
Distance Moved $=80.48\ divisions\times0.5\ mm/division$

Distance Moved $=40.24\ mm$
4. Calculate the Diameter of the Wire:
The diameter of the wire is twice the distance moved by the screw because
the wire is placed between the anvil and the spindle.
Diameter $= 2 \times Distance \ Moved$
Diameter $= 2 \times 40.24 \ mm$
Diameter $= 80.48 \ mm$

Therefore, the diameter of the wire is approximately $80.48 \ mm.$
So, the correct option is B