The pitch of screw gauge is 1mm and its circular scale is divided into 100 divisions. When nothing is put between the zero of main scale is not seen, but when circular scale is rotated by $450^{o}$ the zero of main scale is seen and zero of main scale coincide with the zero of circular scale. When a glass plate is placed between the studs, the circular scale lies between $18^{t h} t o 19^{t h}$ division of main scale and circular scale reds 34 division. Then,

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Error is negative zero error and its magnitude is 1.25 mm

The thickness of the glass plate is 19.59 mm

Error is positive zero error and its magnitude is 1.25 mm

The thickness of the glass plate is 17.09 mm

answer is B, C.

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Detailed Solution

$450^{\circ} = 1 r o t^{n} + \frac{1^{t h}}{4} r o t^{n} = 1 m m + 0.25 m m$

Reading $= 18 + (34 \times 0.01) + (1.25)$

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