CBSE 2026 · Region 2 · Set 2 · Q18 · 2 marks
Calculate the temperature at which the resistance of a conductor becomes $\displaystyle 20$% more than its resistance at $\displaystyle 27$ °C. The value of the temperature coefficient of resistance of the material of conductor is $\displaystyle 2.0 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}$.
Marking-scheme solution
$\displaystyle \mathrm{R}_{2}=\mathrm{R}_{1}\left[1+\alpha\left(t_{2}-t_{1}\right)\right]$
$\displaystyle 1.2 \mathrm{R}_{1}=\mathrm{R}_{1}\left[1+2.0 \times 10^{-4}\left(t_{2}-27\right)\right]$
$\displaystyle 0.2=\left[2.0 \times 10^{-4}\left(t_{2}-27\right)\right]$
$\displaystyle t_{2}=1027^{\circ} \mathrm{C}$
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