A conductor has resistance values of 2.71 Om | vedclass

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(B) The resistance of a conductor at temperature $t$ is given by $R_t = R_{t_0} [1 + \alpha(t - t_0)]$.Given $R_1 = 2.71 \ \Omega$ at $t_1 = 10^{\circ

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$60$

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(B) The resistance of a conductor at temperature $t$ is given by $R_t = R_{t_0} [1 + \alpha(t - t_0)]$.Given $R_1 = 2.71 \ \Omega$ at $t_1 = 10^{\circ}C$ and $R_2 = 3.70 \ \Omega$ at $t_2 = 100^{\circ}C$.Let $R_x = 3.26 \ \Omega$ at temperature $x$.Using the formula: $R_2 = R_1 [1 + \alpha(t_2 - t_1)]$$3.70 = 2.71 [1 + \alpha(100 - 10)]$$3.70 = 2.71 [1 + 90\alpha] \implies 1 + 90\alpha = \frac{3.70}{2.71} \implies 90\alpha = \frac{3.70}{2.71} - 1 = \frac{0.99}{2.71}$.Now,for temperature $x$: $R_x = R_1 [1 + \alpha(x - t_1)]$$3.26 = 2.71 [1 + \alpha(x - 10)]$$\frac{3.26}{2.71} = 1 + \alpha(x - 10) \implies \alpha(x - 10) = \frac{3.26}{2.71} - 1 = \frac{0.55}{2.71}$.Dividing the two expressions for $\alpha$:$\frac{\alpha(x - 10)}{90\alpha} = \frac{0.55 / 2.71}{0.99 / 2.71} \implies \frac{x - 10}{90} = \frac{0.55}{0.99} = \frac{55}{99} = \frac{5}{9}$.$x - 10 = 90 	imes \frac{5}{9} = 50$.$x = 50 + 10 = 60^{\circ}C$.

$A$ conductor has resistance values of $2.71 \ \Omega$ and $3.70 \ \Omega$ at $10^{\circ}C$ and $100^{\circ}C$ respectively. The temperature at which the resistance is $3.26 \ \Omega$ is .......... $^{\circ}C$.

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