JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
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 0.5mm is noticed on the pitch scale. The nature of zero error involved and the least count of the screw gauge, are respectively :
- APositive, 0.1 mm
- BPositive, 0.1 m
- CPositive, 10 m
- DNegative, 2 m
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Correct answer: C
- Find the pitch of the screw gauge
One complete rotation causes a linear displacement of on the pitch scale.
So, the pitch is:
- Find the least count
The circular scale has divisions.
Least count is given by:
Thus,
Now convert to micrometre:
So, the least count is .
- Determine the nature of zero error
The statement says:
the circular scale is 4 units ahead of the pitch scale marking, prior to use.
When the circular scale zero is ahead of the reference line on the pitch scale, the screw gauge shows a reading even when it should read zero. This corresponds to a positive zero error.
Thus, the zero error is positive.
- Check options
- A: Positive, ❌
- B: Positive, ❌
- C: Positive, ✅
- D: Negative, ❌
Hence, the correct option is C.
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