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JEE Mains · Physics · STD 12 - Current electricity
  • JEE Mains 2024
  • Hard
  • Numerical

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

Pick an option to check your answer.
Verified Solution

Answer & Solution

Correct Answer

(D) \(748\)

Step-by-step Solution

Detailed explanation

\(\mathrm{R}=\mathrm{R}_0(1+\alpha \Delta \mathrm{T})\)

\(\frac{\Delta \mathrm{R}}{\mathrm{R}_0}=\alpha \Delta \mathrm{T}\)

Case-I

\(0{ }^{\circ} \mathrm{C} \rightarrow 100^{\circ} \mathrm{C}\)

\(\frac{10.2-10}{10}=\alpha(100-0)\)

Case-II

\(0^{\circ} \mathrm{C} \rightarrow t^{\circ} \mathrm{C}\)

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

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

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

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