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The least count, of a screw gauge, having 100 divisions, on its circular scale, equals 0.0005 cm. When a thin sheet in just held between its jaws, the main scale reads 0.65 mm and the 43rd division, of the circular scale, coincides with the ‘reference line’. The student, doing the experiment, (correctly), reports the thickness of the sheet as 0.0850 cm. The pitch and the ‘zero error’ of this screw gauge, are, respectively, equal to (1) 0.5 mm and (+ 0.0015) cm (2) 0.05 mm and (– 0.0015) cm (3) 0.5 mm and (– 0.0015) cm (4) 0.05 mm and (+ 0.0015) cm |
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Answer» Correct option: (1) 0.5 mm and (+ 0.0015) cm Explanation: We have L.C = Pitch/100 ∴ Pitch = 0.0005 × 100 cm = 0.05 cm = 0.5 mm Total observed reading = (0.65 mm) + (43 × 0.0005) cm = 0.0865 cm Correct reading = Observed reading – (zero error) ∴ Zero error = (0.0865 – 0.0850) cm = + 0.0015 cm |
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