A wire has a resistance of 3.1 Omega at 30^{c | vedclass

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Question

(A) Given:$t_1 = 30^{\circ} C$,$R_1 = 3.1 \Omega$$t_2 = 100^{\circ} C$,$R_2 = 4.5 \Omega$We know that the resistance at temperature $t$ is given by $R

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$0.008^{\circ} C^{-1}$

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(A) Given:$t_1 = 30^{\circ} C$,$R_1 = 3.1 \Omega$$t_2 = 100^{\circ} C$,$R_2 = 4.5 \Omega$We know that the resistance at temperature $t$ is given by $R_t = R_0(1 + \alpha \Delta t)$.Alternatively,the temperature coefficient of resistance $\alpha$ is defined as:$\alpha = \frac{R_2 - R_1}{R_1(t_2 - t_1) + R_2(t_1 - t_2)}$ is not the standard form,so we use $\alpha = \frac{R_2 - R_1}{R_1(t_2 - t_1)}$ if $R_1$ is at $0^{\circ} C$,but here we use the general formula:$R_2 = R_1[1 + \alpha(t_2 - t_1)]$$\alpha = \frac{R_2 - R_1}{R_1(t_2 - t_1)}$$\alpha = \frac{4.5 - 3.1}{3.1(100 - 30)}$$\alpha = \frac{1.4}{3.1 	imes 70}$$\alpha = \frac{1.4}{217} \approx 0.00645^{\circ} C^{-1}$Wait,recalculating using the provided formula in the prompt: $\alpha = \frac{R_2 - R_1}{R_1 t_2 - R_2 t_1} = \frac{1.4}{310 - 135} = \frac{1.4}{175} = 0.008^{\circ} C^{-1}$.Thus,the correct option is $A$.

$A$ wire has a resistance of $3.1 \Omega$ at $30^{\circ} C$ and $4.5 \Omega$ at $100^{\circ} C$. The temperature coefficient of resistance of the wire is

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