Question

There are 100 divisions on the circular scale of a screw gauge of pitch 1 mm. With no measuring quantity in between the jaws, the zero of the circular scale lies 5 divisions below the reference line. The diameter of a wire is then measured using this screw gauge. It is found the 4 linear scale divisions are clearly visible while 60 divisions on circular scale coincide with the reference line. The diameter of the wire is :

• A. 4.65 mm
• B. 4.55 mm
• C. 4.60 mm
• D. 3.35 mm

Answer 1

The Correct Option is B

The procedure to ascertain the wire's diameter via a screw gauge involves the following stages:

1. Parameter Identification:
   • Screw Gauge Pitch: The pitch is established at 1 mm, signifying that each full rotation of the circular scale advances the screw by 1 mm along the linear scale.
   • Circular Scale Divisions: The circular scale comprises 100 divisions.
   • Zero Error Determination: When the jaws are unengaged, the circular scale's zero mark is positioned 5 divisions below the reference line, indicating a negative zero error of 5 divisions.
2. Calculation of Least Count:
   • The least count (L.C.) is computed as: \( \text{L.C.} = \frac{\text{Pitch}}{\text{Number of divisions on the circular scale}} = \frac{1 \, \text{mm}}{100} = 0.01 \, \text{mm} \).
3. Computation of Observed Diameter:
   • The linear scale reading (L.S.R.) is 4 divisions, equivalent to 4 mm.
   • The circular scale reading (C.S.R.) aligning with the reference line is 60 divisions. This contributes an additional measurement of: \( 60 \times \text{L.C.} = 60 \times 0.01 \, \text{mm} = 0.60 \, \text{mm} \).
   • Consequently, the observed diameter is: \( \text{Observed Diameter} = \text{L.S.R.} + \text{C.S.R.} = 4 \, \text{mm} + 0.60 \, \text{mm} = 4.60 \, \text{mm} \).
4. Zero Error Correction:
   • As the zero error is negative (5 divisions below the reference line), the correction applied is: \( -5 \times \text{L.C.} = -5 \times 0.01 \, \text{mm} = -0.05 \, \text{mm} \).
   • Therefore, the corrected wire diameter is: \( 4.60 \, \text{mm} - 0.05 \, \text{mm} = 4.55 \, \text{mm} \).

Final Result: The wire's diameter measures 4.55 mm.