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
A. `0.34+- 0.02 cm`
B. `0.17 +- 0.02 cm`
C. `0.17 + - 0.01cm`
D. `0.34 +- 0.01 cm`

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1 Answers

`D = (4.23 +- 0.01)cm`
`d = (3.89 +- 0.01)cm`
`Deltat = (D -d)//(2)`
`= ((4.23 +-0.01)-(3.9=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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