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 1 2 mm marks. In complete rotation, the screw advances by 1 2 mm and circular scale has 50 divisions. The reading of the screwgauge is

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

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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 $\frac{1}{2}$ $mm$ marks. In complete rotation, the screw advances by $\frac{1}{2}$ $mm$ and circular scale has $50$ divisions. The reading of the screwgauge is

Options:

Answer:

Solution:

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

Stream:JEESubject:PhysicsTopic:Practical physicsSubtopic:Miscellaneous

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