Physics - Current Electricity Question with Solution | TestHub

$\begin{array}{rl}& \mathrm{R}={\mathrm{R}}_0(1+\alpha \Delta \mathrm{T})\\ & \frac{\Delta \mathrm{R}}{{\mathrm{R}}_0}=\alpha \Delta \mathrm{T}\end{array}$ Case-I $\begin{array}{rl}& 0{}^{\circ }\mathrm{C}\to 100^{\circ }\mathrm{C}\\ & \frac{10.2-10}{10}=\alpha (100-0)\end{array}$ ...(1) Case-II $0^{\circ }\mathrm{C}\to {\mathrm{t}}^{\circ }\mathrm{C}$ $\frac{10.95-10}{10}=\alpha (\mathrm{t}-0)$ ...(2) $\begin{array}{rl}& \Rightarrow \frac{\mathrm{t}}{100}=\frac{0.95}{0.2}=475^{\circ }\mathrm{C}\\ & \mathrm{t}=475+273=748\mathrm{K}\end{array}$