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

Answer: (B)

4.55 mm

Answer 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:
$4.55 \, \text{mm}.$