One main-scale division of a Vernier calliper equals 1 mm and the Vernier scale has 10 divisions. When the jaws touch, the Vernier zero is to the left of the main-scale zero and its 4^th division coincides with a main-scale division. If this calliper measures a wire length as 1 cm, the actual length of the wire is:
- A.0.60 cm
- B.0.96 cm
- C.1.00 cm
- D.1.04 cm
Correct Answer
()
Solution & Explanation
Least count of the vernier: L.C. = (1 main-scale division)/(number of vernier divisions) = 1 mm/10 = 0.1 mm = 0.01 cm. Zero error: with the jaws closed, the vernier zero lies to the LEFT of the main-scale zero and the 4th vernier division coincides — this is a negative zero error. Negative zero error = −(N − coinciding division)×L.C. = −(10 − 4)×0.01 = −6×0.01 = −0.06 cm. True length = observed reading − zero error = 1.00 − (−0.06) = 1.00 + 0.06 = 1.06 cm. This value (1.06 cm) does NOT appear among the options. Trap: a left-shifted vernier zero is a negative error, so the correction is ADDED, giving 1.06 cm. Since 1.06 cm is not listed, none of the printed options is correct.
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