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(B) The resistance of a conductor at temperature $t$ is given by $R_t = R_0(1 + \alpha \Delta t)$,where $R_0$ is the resistance at $0\,^{\circ}C$ and

Vedclass · Accepted Answer

$1100$

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(B) The resistance of a conductor at temperature $t$ is given by $R_t = R_0(1 + \alpha \Delta t)$,where $R_0$ is the resistance at $0\,^{\circ}C$ and $\Delta t$ is the temperature change from $0\,^{\circ}C$.Given: $\alpha = 0.00125\,^{\circ}C^{-1}$,$R_1 = 1\,\Omega$ at $T_1 = 300\,K$ $(t_1 = 27\,^{\circ}C)$,and $R_2 = 2\,\Omega$ at $T_2 = ?$.Using the formula $R_2 = R_1[1 + \alpha(t_2 - t_1)]$:$2 = 1[1 + 0.00125(t_2 - 27)]$$2 - 1 = 0.00125(t_2 - 27)$$1 = 0.00125(t_2 - 27)$$t_2 - 27 = \frac{1}{0.00125} = 800$$t_2 = 800 + 27 = 827\,^{\circ}C$Converting to Kelvin: $T_2 = 827 + 273 = 1100\,K$.

The temperature coefficient of resistance for a wire is $0.00125\,^{\circ}C^{-1}$. At $300\,K$ its resistance is $1\,\Omega$. The temperature at which the resistance becomes $2\,\Omega$ is .......... $K$.

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