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Question

In a vernier caliper, 10 vernier scale divisions are equal to 9 main scale divisions. One main scale division is 1 mm. The device has some zero error as shown in the figure (a). When a cylinder of diameter D is inserted between them, the reading of scale is shown in figure (b). Then, the diameter of the cylinder in cm is

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Solution

1VSD=910×1MSD=0.9 mm
LS=1MSD1VSD=1 mm0.9 mm=0.1 mm=0.01 cm
In figure (a), the 6th VSD is alligned with MSD. Therefore the zero error is 6×LS=0.6 mm=0.06 cm
In figure (b), the 4th VSD is alligned with MSD. Therefore the reading is 3 cm+4×(LS)=3.04 cm
Total reading = Actual reading - Error = 3.04 - 0.06 = 2.98 cm

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