The current flowing through a conductor connected across a source...

By Ohm's law, $V = IR$.

Let the current and resistance at $0 ° C$ be ${i}_{0} \text{ and } {R}_{0}$ respectively.

Let the current and resistance at $100 ° C$ be ${i}_{100} \text{ and } {R}_{100}$ respectively.

By the given condition,

${i}_{0}{R}_{0} = {i}_{100}{R}_{100} ⇒ 2{R}_{0} = 1 . 2{R}_{0} ( 1 + 100\alpha ) ⇒ 1 + 100\alpha = \dfrac{5}{3} ⇒ 50\alpha = \dfrac{1}{3}$

Since the voltage remains constant,

${i}_{50}{R}_{50} = {i}_{0}{R}_{0} ⇒ {i}_{50} = \dfrac{{i}_{0}{R}_{0}}{{R}_{50}} = \dfrac{2 \times {R}_{0}}{{R}_{0}\left(1 + 50\alpha\right)} = \dfrac{2}{1 + \dfrac{1}{3}} = 1 . 5A ⇒ {i}_{50} = 15 \times {10}^{2}mA$