Q 11-01-229JEE MainJEE Main 2018 (15 Apr, Shift 1)Easy
In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of $0.25\ \text{cm}$. There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible error, the thickness of the wire is
Answer: (D) $0.2150\ \text{cm}$
Pitch $= \dfrac{0.25}{5} = 0.05\ \text{cm}$, which is also one main scale division.
Least count $= \dfrac{0.05}{100} = 0.0005\ \text{cm}$.
Reading:
$$4\times0.05 + 30\times0.0005 = 0.20 + 0.015 = 0.2150\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics