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In a screw gauge, least count of the pitch scale is 0.5 mm and there are 50 divisions on its cap. When nothing is put between its jaws, 40th division of circular scale coincides with reference line, with zero of circular scale lying above the reference line. When a plate is placed between the jaws, main scale reads 2 divisions and circular scale reads 20 divisions. Thickness of the plate is

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0.8 mm

1.1 mm

1.3 mm

1.6 mm

answer is C.

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Detailed Solution

Least count of Screw Guage $= \frac{0.5 m m}{50} = 0.01 m m$
$e = − 10 × 0.01 = − 0.1 mm$
It has negative zero error
R=2 x 0.5mm + 20 x 0.01 = 1.2 mm
$∴$ True value = 1.2 + 0.1 = 1.3 mm

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