The resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and voltage are each 3%, what is the percentage error in the resistance?

The resistance R of a wire is given by Ohm's law: R = V/I, where V is the voltage and I is the current.

Let's denote the percentage errors in V and I as ΔV and ΔI respectively. We are given that ΔV = 3% and ΔI = 3%.

We want to find the percentage error in R, which we denote as ΔR. We can use the formula for the propagation of uncertainties:

ΔR/R = √[(ΔV/V)² + (ΔI/I)²]

Since ΔV/V = 3% = 0.03 and ΔI/I = 3% = 0.03, we have:

ΔR/R = √[(0.03)² + (0.03)²] = √(0.0009 + 0.0009) = √0.0018 ≈ 0.0424

To express this as a percentage, we multiply by 100:

ΔR/R ≈ 0.0424 * 100 ≈ 4.24%

The closest answer is 6%, however, this solution uses the formula for combining percentage errors for multiplication/division. A more precise answer would be approximately 4.24%. The provided options do not include this value. It's possible there's an error in the options given or a simplification in the expected solution method.