CBSE 2025 · Region 5 · Set 1 · Q21 · 2 marks
The resistance of a wire at $\displaystyle 25^{\circ} \mathrm{C}$ is $\displaystyle 10 \cdot 0 \Omega$. When heated to $\displaystyle 125^{\circ} \mathrm{C}$, its resistance becomes $\displaystyle 10 \cdot 5 \Omega$. Find (i) the temperature coefficient of resistance of the wire, and (ii) the resistance of the wire at $\displaystyle 425^{\circ} \mathrm{C}$.
Marking-scheme solution
(i)
$\displaystyle R_{2}=R_{1}\left(1+\alpha\left(t_{2}-t_{1}\right)\right)$
$\displaystyle 10.5=10(1+\alpha \times 100)$
$\displaystyle \alpha=5 \times 10^{-4} /{ }^{\circ} \mathrm{C}$
(ii)
$\displaystyle R_{425}=R_{25}(1+\alpha(425-25))$
$\displaystyle =10\left(1+5 \times 10^{-4} \times 400\right)$
$\displaystyle =12 \Omega$
Current ElectricityTemperature Dependence of ResistivityApplyvery_short_answermedium
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