The resistance of a copper wire is given by $\mathrm{R}={\mathrm{R}}_0(1+\mathrm{\alpha }(\mathrm{\theta }-20\left)\right]$ where ${\mathrm{R}}_0=(4.00\pm 0.01)\mathrm{\Omega }$ is the resistance of wire at $\mathrm{\theta }={20}^{\circ }\mathrm{C}$. The value of resistance coefficient of copper is ${\text{(4\hspace{0.17em}\hspace{0.17em}×\hspace{0.17em}\hspace{0.17em}10}}^{-3}\mathrm{\hspace{0.17em}}\pm \mathrm{\hspace{0.17em}\hspace{0.17em}}1\times \mathrm{\hspace{0.17em}\hspace{0.17em}}{10}^{-4})/{\mathrm{\hspace{0.17em}}}^{\circ }\mathrm{C}.$ It is connected with a battery of emf $\text{(10.0\hspace{0.17em}\hspace{0.17em}±\hspace{0.17em}\hspace{0.17em}0.1)V.}$

$$\begin{gathered} \text{ At }\mathrm{\theta }={25}^{\circ }\mathrm{C} \\ \mathrm{R}={\mathrm{R}}_0[1+\mathrm{\alpha }(25-20\left)\right] \\ \begin{array}{l}=4\left[1+4\times {10}^{-3}\times 5\right]\\ =4.08\mathrm{\Omega }\end{array} \\ \frac{\mathrm{\Delta R}}{\mathrm{R}}=\frac{{\mathrm{\Delta R}}_0}{{\mathrm{R}}_0}+\frac{\mathrm{\Delta }[1+\mathrm{\alpha }(\mathrm{\theta }-20\left)\right]}{[1+\mathrm{\alpha }(\mathrm{\theta }-20\left)\right]} \\ \begin{array}{l}=\frac{{\mathrm{\Delta R}}_0}{{\mathrm{R}}_0}+\frac{\left(\mathrm{\Delta \alpha }\right)(\mathrm{\theta }-20)}{[1+\mathrm{\alpha }(\mathrm{\theta }-20\left)\right]}\\ =\frac{0.01}{4.00}+\frac{{10}^{-4}\times 5}{1.02}\\ =0.0025+4.9\times {10}^{-4}\\ \frac{\mathrm{\Delta R}}{\mathrm{R}}=0.0029\\ \mathrm{\Delta R}=0.0118\end{array} \end{gathered}$$
$$\begin{gathered} \mathrm{R}=(4.08\pm 0.02)\mathrm{\Omega } \\ \text{ At }\mathrm{\theta }={50}^{\circ }\mathrm{C} \\ \begin{array}{l}\mathrm{R}=4\\ \text{ At }\mathrm{\theta }={50}^{\circ }\mathrm{C}\end{array} \\ \begin{array}{l}\mathrm{R}=4\left(1+4\times {10}^{-3}\times 30\right)\\ \mathrm{R}=4.48\mathrm{\Omega }\end{array} \\ \begin{array}{l}\frac{\mathrm{\Delta R}}{\mathrm{R}}=\frac{\mathrm{\Delta R}}{{\mathrm{R}}_0}+\frac{\left(\mathrm{\Delta \alpha }\right)(\mathrm{\theta }-20)}{[1+\mathrm{\alpha }(\mathrm{\theta }-20\left)\right]}\\ =\frac{0.01}{4.00}+\frac{{10}^{-4}\times 30}{1.12}\end{array} \end{gathered}$$
$$\begin{gathered} \begin{array}{l}=\left(2.5\times {10}^{-3}\right)+\left(2.67\times {10}^{-3}\right)\\ =5.17\times {10}^{-3}\\ \mathrm{\Delta R}=0.023\\ \mathrm{R}=(4.48\pm 0.03)\mathrm{\Omega }\\ \mathrm{P}=\frac{{\mathrm{V}}^2}{\mathrm{R}}\end{array} \\ \begin{array}{l}\text{ At }\mathrm{\theta }={25}^{\circ }\mathrm{C},\mathrm{P}=\frac{100}{4.08}=24.5\mathrm{W}\\ \frac{\mathrm{\Delta P}}{\mathrm{P}}=\frac{2\mathrm{\Delta V}}{\mathrm{V}}+\frac{\mathrm{\Delta R}}{\mathrm{R}}\\ =2\times \frac{0.1}{10}+0.0029\\ =0.02+0.0029\\ \frac{\mathrm{\Delta P}}{\mathrm{P}}=0.0229\end{array} \end{gathered}$$
$\begin{array}{l}\text{ At }\mathrm{\theta }={50}^{\circ }\mathrm{C},\mathrm{P}=\frac{100}{4.48}=22.32\mathrm{W}\\ \frac{\mathrm{\Delta P}}{\mathrm{P}}=2\times \frac{\mathrm{\Delta V}}{\mathrm{V}}+\frac{\mathrm{\Delta R}}{\mathrm{R}}\\ =2\times \frac{0.1}{10}+5.17\times {10}^{-3}\\ =0.025\end{array}$