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 $0.5\text{mm}$ is noticed on the pitch scale. The nature of zero error involved, and the least count of the screw gauge, are respectively:

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

Answer 1

The correct option is B Positive, $10\mu \text{m}$

Given: No. of division on circular scale of screw gauge $=50$
Pitch $=0.5\text{mm}$
Least count of screw gauge
$=\frac{\text{Pitch}}{\text{No. of division on circular scale}}$
$=\frac{0.5}{50}\text{mm}=1\times {10}^{-5}\text{m}=10\mu \text{m}$