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(C) Given: Resistance at room temperature $R_0 = 60 \ \Omega$ at $T_0 = 27^{\circ} C$.Voltage $V = 220 \ V$ and current $I = 2.75 \ A$.The resistance

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$1694$

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(C) Given: Resistance at room temperature $R_0 = 60 \ \Omega$ at $T_0 = 27^{\circ} C$.Voltage $V = 220 \ V$ and current $I = 2.75 \ A$.The resistance at the final temperature $T$ is given by $R_T = \frac{V}{I} = \frac{220}{2.75} = 80 \ \Omega$.The temperature dependence of resistance is given by $R_T = R_0(1 + \alpha \Delta T)$,where $\Delta T = T - T_0$.Substituting the values: $80 = 60[1 + 2 	imes 10^{-4}(T - 27)]$.$80/60 = 1 + 2 	imes 10^{-4}(T - 27)$.$1.333 - 1 = 2 	imes 10^{-4}(T - 27)$.$0.333 = 2 	imes 10^{-4}(T - 27)$.$T - 27 = \frac{0.333}{2 	imes 10^{-4}} = 1665$.$T = 1665 + 27 = 1692^{\circ} C$.Rounding to the nearest option,the temperature is approximately $1694^{\circ} C$.

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