The figure shows two diagrams of the same screw gauge. In the first case, nothing is kept in its jaw and in the second case a small ball is kept between the jaws for which the diameter is required to

Question

The figure shows two diagrams of the same screw gauge. In the first case, nothing is kept in its jaw and in the second case a small ball is kept between the jaws for which the diameter is required to be measured. The number of circular divisions on the shown screw gauge is $50$ . It moves $0.5 mm$ on the main scale for one complete rotation and the main scale has $\frac{1}{2} mm$ marks. The diameter of the ball is

TestHub · Accepted Answer

Pitch = 0.5 mm L.C = 0.5 50 = 0.01 mm M.S.R = 2 Neck reading = 25 d = 2 × 0.5 + 25 × 0.01 = 1.25 mm Considering zero error Here zero error is = + 5 × 0.01 = + 0.05 mm so actual diameter = d - 0.05 = 1.20 mm

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