Question

The figure shown below is a screw gauge. When the circular scale division matches the main scale at $43$, the $6.5mm$ mark on the main scale is just visible. The main scale has $\frac{1}{2}$ $mm$ marks. In complete rotation, the screw advances by $\frac{1}{2}$ $mm$ and circular scale has $50$ divisions. The reading of the screw gauge is

• A. 6.93 mm
• B. 5.93 mm
• C. 4.93 mm
• D. 7.93 mm

Answer 1

Answer: (A)

Object thickness $=6.5mm+43\frac{1/2mm}{50}=6.93mm$