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 :
- 4.65 mm
- 4.55 mm
- 4.60 mm
- 3.35 mm
The Correct Option is B
Approach Solution - 1
To determine the diameter of the wire using the screw gauge, we need to follow these steps:
- Understanding the Key Parameters:
- Pitch of the screw gauge: The pitch is given as 1 mm, which means one complete rotation of the circular scale moves the screw forwards or backwards by 1 mm on the linear scale.
- Total divisions on the circular scale: There are 100 divisions.
- Zero error: The zero of the circular scale is 5 divisions below the reference line when there is nothing between the jaws. This indicates a negative zero error of 5 divisions.
- Calculate the Least Count of the Screw Gauge:
- The least count (L.C.) can be calculated as follows: \(\text{L.C.} = \frac{\text{Pitch}}{\text{Number of divisions on the circular scale}} = \frac{1 \, \text{mm}}{100} = 0.01 \, \text{mm}\).
- Calculating the Observed Diameter of the Wire:
- The linear scale reading (L.S.R.) shows 4 divisions, which corresponds to 4 mm.
- The circular scale reading (C.S.R.) that coincides with the reference line is 60 divisions. This adds an additional length of: \(60 \times \text{L.C.} = 60 \times 0.01 \, \text{mm} = 0.60 \, \text{mm}\).
- Thus, the observed diameter is: \(\text{Observed Diameter} = \text{L.S.R.} + \text{C.S.R.} = 4 \, \text{mm} + 0.60 \, \text{mm} = 4.60 \, \text{mm}\).
- Correcting for Zero Error:
- The zero error is negative (5 divisions below), so the correction will be: \(-5 \times \text{L.C.} = -5 \times 0.01 \, \text{mm} = -0.05 \, \text{mm}\).
- Thus, the corrected diameter of the wire is: \(4.60 \, \text{mm} - 0.05 \, \text{mm} = 4.55 \, \text{mm}\).
Conclusion: The diameter of the wire is 4.55 mm. Thus, the correct answer is 4.55 mm.
Approach Solution -2
The least count of the screw gauge is:
\[\text{Least count} = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}} = \frac{1 \, \text{mm}}{100} = 0.01 \, \text{mm}.\]
The zero error is given as:
\[\text{Zero error} = +0.05 \, \text{mm}.\]
The reading is calculated as:
\[\text{Reading} = (\text{Linear scale reading}) \times (\text{Pitch}) + (\text{Circular scale reading}) \times (\text{Least count}) - \text{Zero error}.\]
Substitute:
\[\text{Reading} = (4 \times 1) \, \text{mm} + (60 \times 0.01) \, \text{mm} - 0.05 \, \text{mm}.\]
Simplify:
\[\text{Reading} = 4.00 \, \text{mm} + 0.60 \, \text{mm} - 0.05 \, \text{mm} = 4.55 \, \text{mm}.\]
Thus, the diameter of the wire is:
\[4.55 \, \text{mm}.\]
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