Resistance of a wire at 0 ° C , 100 ° C and t ° C is found to be 10 Ω, 10.2 Ω and 10.95 Ω respectively. The temperature t in Kelvin scale is .

Physics · Thermal Properties of Matter

JEE Main 2024 — 8 April, Shift 1 — Question 51

Resistance of a wire at $0^{\circ} \mathrm{C}, 100^{\circ} \mathrm{C}$ and $\mathrm{t}{ }^{\circ} \mathrm{C}$ is found to be $10 \Omega, 10.2 \Omega$ and $10.95 \Omega$ respectively. The temperature $t$ in Kelvin scale is $\qquad$ .

Answer: 748

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{R}=\mathrm{R}_{0}(1+\alpha \Delta \mathrm{T})$

\frac{\Delta \mathrm{R}}{\mathrm{R}_{0}}=\alpha \Delta \mathrm{T}

Case-I

$0^{\circ} \mathrm{C} \rightarrow 100^{\circ} \mathrm{C}$

\begin{gathered} \frac{10.2-10}{10}=\alpha(100-0)….(1) \end{gathered}

Case-II

$0^{\circ} \mathrm{C} \rightarrow \mathrm{t}^{\circ} \mathrm{C}$

$\frac{10.95-10}{10}=\alpha(\mathrm{t}-0)$

$\Rightarrow \frac{\mathrm{t}}{100}=\frac{0.95}{0.2}=475^{\circ} \mathrm{C}$

$\mathrm{t}=475+273=748 \mathrm{~K}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 8 April, Shift 1
Subject: Physics
Chapter: Thermal Properties of Matter
Topic: Thermal Expansion of Solids and its Applications