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Thermal Properties of Matter

JEE Main level, numerical answer, medium

17 of 126

Question 17 of 126

At room temperature $\left(27^{\circ} \mathrm{C}\right)$ , the resistance of a heating element is $50 \Omega$ . The temperature coefficient of the material is $2.4 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}$ . The temperature of the element, when its resistance is $62 \Omega$ , is _____ ${ }^{\circ} \mathrm{C}$ .

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Answer: 1027

\begin{aligned} & \mathrm{R}=\mathrm{R}_0(1+\alpha \Delta \mathrm{T}) \\ & 62=50\left[1+2.4 \times 10^{-4} \Delta \mathrm{T}\right] \\ & \Delta \mathrm{T}=1000^{\circ} \mathrm{C} \\ & \Rightarrow \mathrm{T}-27^{\circ}=1000^{\circ} \mathrm{C} \\ & \mathrm{T}=1027^{\circ} \mathrm{C}\end{aligned}

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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