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(D) Since the conductor is connected across the same source,the voltage $V$ remains constant. Thus,$V = i_0 R_0 = i_{100} R_{100} = i_{50} R_{50}$.Giv

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$15$

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(D) Since the conductor is connected across the same source,the voltage $V$ remains constant. Thus,$V = i_0 R_0 = i_{100} R_{100} = i_{50} R_{50}$.Given $i_0 = 2\,A$ and $i_{100} = 1.2\,A$.Using the resistance-temperature relation $R_t = R_0(1 + \alpha t)$:$2 R_0 = 1.2 R_0(1 + 100 \alpha)$$1 + 100 \alpha = \frac{2}{1.2} = \frac{20}{12} = \frac{5}{3}$$100 \alpha = \frac{5}{3} - 1 = \frac{2}{3} \Rightarrow 50 \alpha = \frac{1}{3}$.Now,for $t = 50^{\circ}C$:$i_{50} = \frac{V}{R_{50}} = \frac{i_0 R_0}{R_0(1 + 50 \alpha)} = \frac{2}{1 + \frac{1}{3}} = \frac{2}{4/3} = \frac{6}{4} = 1.5\,A$.Converting to $mA$: $1.5\,A = 1500\,mA = 15 	imes 10^2\,mA$.

The current flowing through a conductor connected across a source is $2\,A$ and $1.2\,A$ at $0^{\circ}C$ and $100^{\circ}C$ respectively. The current flowing through the conductor at $50^{\circ}C$ will be $......\times 10^2\,mA$.

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