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.
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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