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Q.

In a screw gauge 5 complete rotations of circular scale gives 1.5 mm reading on linear scale. If the circular scale has 50 divisions then the least count of screw gauge is $x \times 10^{y} (m m)$ . Then $x - y =$

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Detailed Solution

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Pitch = $\frac{1.5}{5} = 0.3 m m$
Least count $= \frac{0.3}{50} = 0.006 m m = 6 \times 10^{- 3} m m$

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