Question

A screw gauge has $50$ divisions on its circular scale. The circular scale is $4 $ units ahead of the pitch scale marking, prior to use. Upon one complete rotation of the circular scale, a displacement of $0.5\, mm$ is noticed on the pitch scale. The nature of zero error involved, and the least count of the screw gauge, are respectively :

• A. Negative, $2\, \mu m$
• B. Positive, $10\, \mu m$
• C. Positive, $0.1\, \mu m$
• D. Positive, $0.1\, mm$

Answer 1

The Correct Option is B

To solve this problem, we need to determine the nature of the zero error and the least count of a screw gauge given the specifics of its scales.

1. Understand the given data:
   • The screw gauge has 50 divisions on its circular scale.
   • Upon one complete rotation of the circular scale, a displacement of 0.5 \, \text{mm} is noticed on the pitch scale.
   • The circular scale is 4 units ahead of the pitch scale marking, prior to use.
2. Calculate the least count of the screw gauge:
   • The least count is calculated by dividing the pitch by the number of divisions on the circular scale.
   • Pitch of the screw gauge = Displacement during one complete rotation = 0.5 \, \text{mm}
   • Number of divisions on circular scale = 50
   • Thus, Least Count (LC) = \frac{0.5 \, \text{mm}}{50} = 0.01 \, \text{mm}
   • Converting to micrometers: 0.01 \, \text{mm} = 10 \, \mu m
3. Determine the zero error:
   • The circular scale is 4 divisions ahead of zero, indicating a positive zero error.
   • Positive zero error occurs when the zero mark on the circular scale is ahead of the baseline or zero mark on the pitch scale.
4. Conclusion:
   • Therefore, the nature of the zero error is positive, and the least count of the screw gauge is 10 \, \mu m.

Thus, the correct answer is that the nature of the zero error is "Positive" and the least count is 10 \, \mu m.