The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0 . 5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its cir

Question

The pitch and the number of divisions, on the circular scale, for a given screw gauge are $0.5 mm$ and $100$ respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies $3$ divisions below the mean line.

The readings of the main scale and the circular scale, for a thin sheet, are $5 .5 mm$ and $48$ respectively, the thickness of this sheet is:

Options:

Answer:

C

Solution:

least count $= \frac{0.5 m m}{100} = 0.005 m m$

Reading $= 5.5 m m + (48 - 3) \times 0.005 m m$

$= 5.725 m m .$

TestHub · Accepted Answer

least count = 0.5 m m 100 = 0.005 m m Reading = 5.5 m m + 48 - 3 × 0.005 m m = 5.725 m m .

Physics - Error Question with Solution | TestHub

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