JEE Main 2019 Physics question
The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line. The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is:
Answer: (C)
Solution
LC = 0.005 mm; reading 5.5 + 48 × 0.005 = 5.740 mm; zero error −0.015 mm ⇒ 5.755 mm.