The external and internal diameters of a hollow cylinder are measured to be (4.23±0.01) cm and (3.89 ± 0.01) cm. The thickness of the wall of the cylinder is:

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(0.34±0.02)cm

(0.17±0.02)cm

(0.17±0.01)cm

(0.34±0.01)cm

answer is C.

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Detailed Solution

D=(4.23±0.01)cmd=(3.89±0.01)cmΔt=(D−d)2 =(4.23±0.01)−(3.89±0.01)2 =(4.23−3.89)±(0.01+0.01)2 =(0.34±0.02)2cm =(0.17±0.01)cm

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