CBSE 2025, Set 1

Question

The resistance of a wire at 25°C is 10·0 $\Omega$. When heated to 125°C, its resistance becomes 10·5 $\Omega$. Find (i) the temperature coefficient of resistance of the wire, and (ii) the resistance of the wire at 425°C.

Edvaya · Accepted Answer

The correct answer is Subjective. Using the relation: $R_2 = R_1[1 + \alpha(T_2 - T_1)]$ **(i) Temperature coefficient $\alpha$:** $$10\cdot5 = 10\cdot0 	imes [1 + \alpha(125 - 25)]$$ $$10\cdot5 = 10\cdot0 	imes [1 + 100\alpha]$$ $$1 + 100\alpha = 1\cdot05 \implies 100\alpha = 0\cdot05$$ $$\boxed{\alpha = 5 	imes 10^{-4} 	ext{ /°C}}$$ **(ii) Resistance at 425°C:** $$R_{425} = R_{25}[1 + \alpha(425 - 25)] = 10\cdot0 	imes [1 + 5 	imes 10^{-4} 	imes 400]$$ $$= 10\cdot0 	imes [1 + 0\cdot2] = 10\cdot0 	imes 1\cdot2 = \boxed{12 	ext{ }\Omega}$$ > **Note:** The temperature coefficient $\alpha$ for metals is positive (resistance increases with temperature), as higher temperature increases lattice vibrations and scattering of conduction electrons. For semiconductors, $\alpha$ is negative.

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