Current Electricity

JEE Main level, numerical answer, medium

48 of 380

Question 48 of 380

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 _________

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Answer: 748

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∘C0^{\circ} \mathrm{C} \rightarrow \mathrm{t}^{\circ} \mathrm{C}
10.95−1010=α(t−0)\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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