The resistance of an iron wire is 10,Omega an | vedclass

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(A) The resistance of a conductor at temperature $t$ is given by $R_t = R_0(1 + \alpha \Delta t)$.Given $R_1 = 10\,\Omega$ at $t_1 = 20\,^{\circ}C$ an

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$20$

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(A) The resistance of a conductor at temperature $t$ is given by $R_t = R_0(1 + \alpha \Delta t)$.Given $R_1 = 10\,\Omega$ at $t_1 = 20\,^{\circ}C$ and $\alpha = 5 	imes 10^{-3}\,^{\circ}C^{-1}$.The resistance at $t_2 = 120\,^{\circ}C$ is $R_2 = R_1[1 + \alpha(t_2 - t_1)]$.$R_2 = 10[1 + 5 	imes 10^{-3} 	imes (120 - 20)] = 10[1 + 5 	imes 10^{-3} 	imes 100] = 10[1 + 0.5] = 15\,\Omega$.Since the potential difference $V$ is constant,$V = I_1 R_1 = I_2 R_2$.Therefore,$I_2 = I_1 	imes (R_1 / R_2) = 30\,mA 	imes (10 / 15) = 30 	imes (2/3) = 20\,mA$.

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