The figure shown below is a screwgauge. When the circular scale division matches the main scale at 43, the $6 . 5 m m$ mark on the main scale is just visible. The main scale has $\dfrac{1}{2}mm$ marks. In complete rotation, the screw advances by $\dfrac{1}{2}mm$ and circular scale has $50$ divisions. The reading of the screwgauge is

Question

The figure shown below is a screwgauge. When the circular scale division matches the main scale at 43, the

6 . 5 m m

mark on the main scale is just visible. The main scale has

\dfrac{1}{2}mm

marks. In complete rotation, the screw advances by

\dfrac{1}{2}mm

and circular scale has

50

divisions. The reading of the screwgauge is

The figure shown below is a screwgauge. When the circular scale division matches the main scale at 43, the 6 . 5 m m

The figure shown below is a screwgauge. When the circular scale division matches the main scale at 43, the 6 . 5 m m

Quantrex Academy · Accepted Answer

Correct answer: A. 6 . 93mm

Object thickness = 6.5 mm + 4 3 (1 / 2 mm)/(5 0) = 6.93 mm

The figure shown below is a screwgauge. When the circular scale...

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