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Find the percentage error in specific resistance given by `rho = (pir^2R)/(l)` where r is the radius having value `(0.2+-0.02)`cm, R is the resistance of `(60+-2) ohm` and l is the length of `(150+-0.1)`cm. |
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Answer» Correct Answer - B::C::D `(Delta rho)/(p) xx 100=[2(Deltar)/(r)+(DeltaR)/(R)+(DeltaI)/(I)]xx100 ` ` =[(2xx0.002)/(0.2)+(2)/(60)+(0.1/150)]xx100 ` `23.4%` |
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