The resistance of a wire of iron is $10\, ohms$ and temp. coefficient of resistivity is $5 \times {10^{ - 3}}\,^oC$. At $20\,^oC$ 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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$\frac{{{R_1}}}{{{R_2}}} = \frac{{(1 + \alpha {t_1})}}{{(1 + \alpha {t_2})}}$
$\frac{{10}}{{{R_2}}} = \frac{{(1 + 5 \times {{10}^{ - 3}} \times 20)}}{{(1 + 5 \times {{10}^{ - 3}} \times 120)}}$
${R_2} \approx 15\,\Omega $
Also $\frac{{{i_1}}}{{{i_2}}} = \frac{{{R_2}}}{{{R_1}}}$
$\frac{{30}}{{{i_2}}} = \frac{{15}}{{10}}$
${i_2} = 20\,mA$
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