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If measured value of resistance `R = 1.05Omega` wire diameter `d = 0.60 mm` and length `l = 75.3 cm` If maximum error in resistance measurment is `0.01 Omega` and least count of diameter and length measuring device are `0.01 mm` and `0.1cm` respectively then find max permissible error in resistiviry `rho=(R((pid^(2))/(4)))/(l)` . |
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Answer» `((Deltarho)/(rho))_(max)=(DeltaR)/(R)+2(Deltad)/(d)+(Deltal)/(l)` `DeltaR = 0.01 Omega` `Deltad = 0.01 mm` (least count) `Deltal = 0.1 cm` (least count) `((Deltarho)/(rho))_(max)=((0.01Omega)/(1.05Omega)+2(0.01mm)/(0.60mm)+(0.1cm)/(75.3cm))xx100=4.3%` . |
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