Mathematics in Physics

JEE Main level, single correct, medium

71 of 121

Question 71 of 121

The resistance R=VIR=\dfrac{V}{I}, where V=(200±5)VV=(200\pm 5)V and I=(20±0.2)AI=(20\pm 0.2)A, the percentage error in the measurement of RR is :

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Answer: (A) 3.53.5%

Given that, R=VIR=\dfrac{V}{I}
According to error analysis
dRR=dVV+dII⇒dRR=5200+0.220⇒dRR=7200\dfrac{dR}{R}=\dfrac{dV}{V}+\dfrac{dI}{I} \Rightarrow \dfrac{dR}{R}=\dfrac{5}{200}+\dfrac{0.2}{20} \Rightarrow \dfrac{dR}{R}=\dfrac{7}{200}
Thus, the percentage error can be calculated as follows:
%\text{error}=\dfrac{dR}{R}\times 100 =\dfrac{7}{200}\times 100 =3.5%

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