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(B) The resistance of a conductor at a temperature $T$ is given by the formula: $R = R_0(1 + \alpha \Delta T)$.Here,$R_0 = 50 \Omega$ at $T_0 = 27^{\c

Vedclass · Accepted Answer

$1027$

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(B) The resistance of a conductor at a temperature $T$ is given by the formula: $R = R_0(1 + \alpha \Delta T)$.Here,$R_0 = 50 \Omega$ at $T_0 = 27^{\circ} C$,$R = 62 \Omega$,and $\alpha = 2.4 	imes 10^{-4} { }^{\circ} C^{-1}$.Substituting the values into the equation:$62 = 50(1 + 2.4 	imes 10^{-4} \Delta T)$$1.24 = 1 + 2.4 	imes 10^{-4} \Delta T$$0.24 = 2.4 	imes 10^{-4} \Delta T$$\Delta T = \frac{0.24}{2.4 	imes 10^{-4}} = 1000^{\circ} C$.Since $\Delta T = T - T_0$,we have $T = T_0 + \Delta T = 27^{\circ} C + 1000^{\circ} C = 1027^{\circ} C$.

At room temperature $(27^{\circ} C)$,the resistance of a heating element is $50 \Omega$. The temperature coefficient of the material is $2.4 \times 10^{-4} { }^{\circ} C^{-1}$. The temperature of the element,when its resistance is $62 \Omega$,is $\qquad$ ${ }^{\circ} C$.

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