Pls ans following urgently dear experts:
Q. A micrometer screw gauge has a negative error of 0.008 cm. While measuring the diameter of a wire, the reading on the main scale is 3 divisions and the 24th circular scale division coincides with the baseline. If the number of divisions on the main scale are 20 and circular scale has 100 divisions, calculate the:
(a) pitch of the screw gauge.            [Ans. 0.1 cm]
(b) least count of the screw gauge.  [Ans. 0.001 cm]
(c) observed diameter of the wire.   [Ans. 0.174 cml
(d) corrected diameter of the wire.  [Ans. 0.182 cm]

Dear Student,
Please find the solution to the asked query

Pitch = 0.1cm or 1mmLeast count = pitchTotal dividsion on the circular scale = 0.1100 = 0.001 cmObserved reading = Linear Scale Reading + no. of divisions on the circular scale × least countObserved reading = 3mm +24×0.001 = 3.024 mmZero Error = - 8×0.001 = -0.008 mmActual reading = Observed reading - zero errorActual Reading = 3.024 -(-0.008) = 3.032 mm


Regards

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