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The external and internal diameters of a hollow cylinder are measured to be `(4.23 pm 0.01)cm ` and `(3.89pm0.01) cm`. The thickness of the wall of the cylinder isA. `(0.34 pm 0.02)` cmB. `(0.17 pm 0.02)` cmC. `(0.17 pm 0.01)` cmD. `(0.34 pm 0.01)` cm |
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Answer» Correct Answer - A Thickness of cylinder can be written as `t = t pm Delta t` `t= r_2 - r_1 = 4.23 - 3.89 = 0.34` `Delta t = Delta r_1 + Delta r_2 = 0.01 + 0.01 = 0.02` ` t (0.34 pm 0.02)` cm |
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