The resistance of a wire of iron is 10 ohms and temp. coefficient of resistivity is $5 \times 10^{- 3} /^{\circ} C . At 20^{\circ} C$ it carries 30 milliamperes of current. Keeping constant potential difference between its ends, the temperature of the wire is raised to 120^oC. The current in milliamperes that flows in the wire is

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Detailed Solution

$\frac{R_{1}}{R_{2}} = \frac{(1 + {αt}_{1})}{(1 + α t_{2})} \Rightarrow \frac{10}{R_{2}} = \frac{(1 + 5 \times 10^{- 3} \times 20)}{(1 + 5 \times 10^{- 3} \times 120)} \Rightarrow R_{2} \approx 15 Ω Also \frac{i_{1}}{i_{2}} = \frac{R_{2}}{R_{1}} \Rightarrow \frac{30}{i_{2}} = \frac{15}{10} \Rightarrow i_{2} = 20 mA$

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