A student measures the external diameter of a hollow tube using a vernier calipers for which 25 VSD = 24 MSD and 1 MSD = 1 mm. The main scale reading is 26 mm and the 18th vernier division coincides. The instrument has a positive zero error of +0.08 mm. The corrected internal diameter of the tube, measured separately, is 21.84 mm. The wall thickness is:
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Question

A student measures the external diameter of a hollow tube using a vernier calipers for which 25 VSD = 24 MSD and 1 MSD = 1 mm. The main scale reading is 26 mm and the 18th vernier division coincides. The instrument has a positive zero error of +0.08 mm. The corrected internal diameter of the tube, measured separately, is 21.84 mm. The wall thickness is:

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G1 Physics · Accepted Answer

LC = 1/25 mm = 0.04 mm. Observed external diameter = 26 + 18(0.04) = 26.72 mm. Corrected external diameter = 26.72 − 0.08 = 26.64 mm. Wall thickness = (26.64 − 21.84)/2 = 2.40 mm.