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(B) The resistance of a conductor at temperature $T$ is given by $R_T = R_0(1 + \alpha T)$,where $R_0$ is the resistance at $0\,^{\circ}	ext{C}$.Give

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$190\,^{\circ}	ext{C}$

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(B) The resistance of a conductor at temperature $T$ is given by $R_T = R_0(1 + \alpha T)$,where $R_0$ is the resistance at $0\,^{\circ}	ext{C}$.Given $\alpha = 0.00125\,^{\circ}	ext{C}^{-1}$.At $T_1 = 25\,^{\circ}	ext{C}$,$R_1 = 1\,\Omega$. So,$1 = R_0(1 + 0.00125 	imes 25) = R_0(1 + 0.03125) = R_0(1.03125)$.Thus,$R_0 = \frac{1}{1.03125} \approx 0.9697\,\Omega$.We want to find $T_2$ such that $R_2 = 1.2\,\Omega$.Using $R_2 = R_0(1 + \alpha T_2)$,we have $1.2 = R_0(1 + 0.00125 T_2)$.Substituting $R_0$: $1.2 = \frac{1}{1.03125} (1 + 0.00125 T_2)$.$1.2 	imes 1.03125 = 1 + 0.00125 T_2$.$1.2375 = 1 + 0.00125 T_2$.$0.2375 = 0.00125 T_2$.$T_2 = \frac{0.2375}{0.00125} = 190\,^{\circ}	ext{C}$.

The temperature coefficient of resistance at $0\,^{\circ}\text{C}$ is $\alpha = 0.00125\,^{\circ}\text{C}^{-1}$. At a temperature of $25\,^{\circ}\text{C}$,the resistance of a conductor is $1\,\Omega$. Find the temperature at which the resistance becomes $1.2\,\Omega$.

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