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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 isA. `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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Answer» `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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