JEE Main 2019 Physics question
The least count of the main scale of a screw gauge is 1 mm. The minimum number of divisions on its circular scale required to measure 5 μm diameter of a wire is:
Answer: (C)
Solution
LC = 1 mm/N = 5 μm ⇒ N = 200.
The least count of the main scale of a screw gauge is 1 mm. The minimum number of divisions on its circular scale required to measure 5 μm diameter of a wire is:
LC = 1 mm/N = 5 μm ⇒ N = 200.