Q.
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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a
(0.34±0.02)cm
b
(0.17±0.02)cm
c
(0.17±0.01)cm
d
(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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