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The resistance $$R = \frac{V}{I}$$, where $$V = (200 \pm 5)$$ V and $$I = (20 \pm 0.2)$$ A, the percentage error in the measurement of $$R$$ is :
$$R = V/I$$. Percentage error in R:
$$\frac{\Delta R}{R} \times 100 = \frac{\Delta V}{V} \times 100 + \frac{\Delta I}{I} \times 100$$
$$= \frac{5}{200} \times 100 + \frac{0.2}{20} \times 100 = 2.5\% + 1\% = 3.5\%$$
The answer is $$3.5\%$$, which corresponds to Option (1).
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