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. Negative, $2 \mu \mathrm{m}$

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

  3. Positive, $0.1 \mu \mathrm{m}$

  4. Positive, $0.1 \mathrm{~mm}$


Correct Option: , 2

Solution:

Least count of screw gauge

$\frac{\text { Pitch }}{\text { sion on circular scale }}$

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

$=10 \mu \mathrm{m}$

Zero error in positive

Ans. (2)

 

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