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  • A wire of unknown diameter is measured using a screw gauge. The screw gauge has  divisions on its circular scale, and

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


Answers (1)

best_answer

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
reading:

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

2. Calculate the Actual Reading:
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 \timesPitch
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

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