The specific resistance σ of a thin wire of radius r cm, resistance R Ω and length L cm is given by σ = (πr^2R)/L. If r = 0.26 +/- 0.02 cm, R = 32 +/- 1 Ω and L = 78.00 +/- 0.01 cm, find the % error in σ.
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Explanation:
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Answer:
The percentage error in the specific resistance is 18.5%.
Explanation:
Given that,
The specific resistance
Radius r = 0.26±0.02 cm
Length l =78.00±0.01 cm
Resistance R =32±1 ohm
The percentage error is defined as,
The percentage error in the specific resistance is
Hence, The percentage error in the specific resistance is 18.5%.
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