Measuring Instruments & Errors — A/L Physics

How do I practise Measuring Instruments & Errors?

1 question with a worked solution

1. A micrometer screw gauge has a pitch of 0.50 mm and 50 circular scale divisions. When closed without any object, the circular scale zero is 2 divisions above the reference datum line (-0.02 mm negative zero error). In measuring the diameter of a wire, the main scale reads 2.50 mm and the circular scale reads 36 divisions. What is the corrected diameter of the wire?
   1. A) 2.84 mm
   2. B) 2.88 mm
   3. C) 2.90 mm
   4. D) 2.86 mm
   Check my answer

   Correct answer: B) 2.88 mm

   Least count = Pitch / divisions = 0.50 mm / 50 = 0.01 mm. Observed reading = Main scale + (circular division × least count) = 2.50 mm + (36 × 0.01 mm) = 2.86 mm. Corrected reading = Observed reading - Zero error = 2.86 mm - (-0.02 mm) = 2.88 mm.

   Show worked solution

   Step 1: Calculate least count (LC) = 0.50 mm / 50 = 0.01 mm. Step 2: Observed reading = 2.50 mm + 36 × 0.01 mm = 2.86 mm. Step 3: Corrected reading = Observed - (Zero error) = 2.86 mm - (-0.02 mm) = 2.88 mm. Answer: 2.88 mm.
   Modelled on: the A/L Physics examination pattern

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Where does Measuring Instruments & Errors fit in the syllabus?

Unit 1.4 · Year 1 · Term 1