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A screw gauge with a pitch of 0.5  mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of aluminum. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the 45 t h division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness (in m m ) of the sheet if the mai

Correct answer

0.85

Step-by-step solution

Least count, L C = 0 . 5 50 = 0 . 01   mm Zero error, Z E = 50 - 45 × 0 . 01 = - 0 . 05   mm Main Scale Reading M S R = 0.5   mm Thickness, t = 0 . 5   mm   + 30 × 0 . 01   mm   -   - 0 . 05   mm t = 0 . 85   mm

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