A screw gauge has 50 divisions on its circular scale.
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 $0.5 \mathrm{~mm}$ is noticed on the pitch scale. The nature of zero error involved, and the least count of the screw gauge, are respectively:

  1. (1) Negative, $2 \mu \mathrm{m}$

  2. (2) Positive, $10 \mu \mathrm{m}$

  3. (3) Positive, $0.1 \mathrm{~mm}$

  4. (4) Positive, $0.1 \mu \mathrm{m}$


Correct Option: 2

Solution:

(2) Given : No. of division on circular scale of screw gauge

$=50$

Pitch $=0.5 \mathrm{~mm}$

Least count of screw gauge

$=\frac{\text { Pitch }}{\text { No. of division on circular scale }}$

$=\frac{0.5}{50} \mathrm{~mm}=1 \times 10^{5} \mathrm{~m}=10 \mu \mathrm{m}$

And nature of zero error is positive.

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