$T_1$ = 27.5 °C , $R_1$ = 2.1 Ω , $T_2$ = 100 °C and $R_2$ = 2.7 Ω Using the relation, $R_2$ = $R_1$ [1 + α $(T_2-T_1)$] , we have Temperature coefficient of resistivity of silver, α = $\frac{R_2-R_1}{R_1(T_2-T_1)}$ = $\frac{2.7-2.1}{2.1\times(100-27.5)}$ = $\frac{0.6}{152.25}$ = 3.94 × $10^{-3}$ or 0.0039 $^{\circ}\mathrm{C}^{-1}$