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A student measures the resistance, diameter, and length of a wire to determine its electrical resistivity. The measured values are: Resistance R = 25.0 0.5 Diameter d = 1.00 0.02 mm Length l = 100.0 0.5 cm The calculated maximum percentage error in the resistivity of the wire is:

Options

  1. A4.5 %
  2. B3.5 %
  3. C5.5 %
  4. D6.5 %

Correct answer

D. 6.5 %

Step-by-step solution

The resistivity of a wire is given by the formula: = R A l = R d^2 4 l The maximum percentage error in resistivity is given by the sum of the percentage errors of the individual quantities, taking into account their powers: 100 = ( R R + 2 d d + l l ) 100 Calculating the individual percentage errors: Percentage error in R = 0.5 25.0 100 = 2 % Percentage error in d = 0.02 1.00 100 = 2 % Percentage error in l = 0.5 100.0 100 = 0.5 % Substituting these values into the error equation: 100 = 2 % + 2(2 %) + 0.5 % 100 = 2

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