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JEE Main 20248 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
8 April
Shift
Morning Shift
Subject
Physics
Chapter
Current Electricity

JEE Main 2024 · 8 April · Morning Shift · Physics · Current Electricity

Resistance of a wire at 0^ C, 100^ C and t^ C is found to be 10 ,...

Resistance of a wire at 0∘C,100∘C0^{\circ} \mathrm{C}, 100^{\circ} \mathrm{C} and t∘Ct^{\circ} \mathrm{C} is found to be 10Ω,10.2Ω10 \Omega, 10.2 \Omega and 10.95Ω10.95 \Omega respectively. The temperature tt in Kelvin scale is _________

Correct answer

0

Step-by-step solution

R=R0(1+αΔT)ΔRR0=αΔT\begin{aligned} & \mathrm{R}=\mathrm{R}_0(1+\alpha \Delta \mathrm{T}) \\ & \frac{\Delta \mathrm{R}}{\mathrm{R}_0}=\alpha \Delta \mathrm{T}\end{aligned} Case-I 0∘C→100∘C10.2−1010=α(100−0)\begin{aligned} & 0{ }^{\circ} \mathrm{C} \rightarrow 100^{\circ} \mathrm{C} \\ & \frac{10.2-10}{10}=\alpha(100-0)\end{aligned} ...(1) Case-II 0∘C→t∘C10.95−1010=α(t−0)0^{\circ} \mathrm{C} \rightarrow \mathrm{t}^{\circ} \mathrm{C} \frac{10.95-10}{10}=\alpha(\mathrm{t}-0) ...(2) ⇒t100=0.950.2=475∘Ct=475+273=748 K\begin{aligned} & \Rightarrow \frac{\mathrm{t}}{100}=\frac{0.95}{0.2}=475^{\circ} \mathrm{C} \\ & \mathrm{t}=475+273=748 \mathrm{~K}\end{aligned}
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