Q 11-01-196JEE MainJEE Main 2020 (9 Jan, Shift 1)Easy
If the screw on a screw-gauge is given six rotations, it moves by $3$ mm on the main scale. If there are $50$ divisions on the circular scale the least count of the screw gauge is:
Answer: (A) $0.001$ cm
Pitch $= \dfrac{3\ \text{mm}}{6} = 0.5$ mm.
Least count $= \dfrac{\text{pitch}}{\text{circular divisions}} = \dfrac{0.5}{50} = 0.01$ mm $= 0.001$ cm.
Solution by Sreeraj P, M.Sc Physics