The resistance of a wire at 30^{circ}C is 10, | vedclass

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(B) Given: Initial temperature $T_1 = 30^{\circ}C$,Initial resistance $R_1 = 10\, \Omega$,Temperature coefficient $\alpha = 0.002\, ^{\circ}C^{-1}$.We

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$83$

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(B) Given: Initial temperature $T_1 = 30^{\circ}C$,Initial resistance $R_1 = 10\, \Omega$,Temperature coefficient $\alpha = 0.002\, ^{\circ}C^{-1}$.We want to increase the resistance by $10\%$,so the new resistance $R_2 = R_1 + 0.10 R_1 = 1.10 R_1 = 11\, \Omega$.The formula for resistance at temperature $T$ is $R = R_0(1 + \alpha T)$.For $T_1 = 30^{\circ}C$: $R_1 = R_0(1 + 30\alpha) = 10\, \Omega$.For $T_2$: $R_2 = R_0(1 + \alpha T_2) = 11\, \Omega$.Dividing the two equations: $\frac{R_2}{R_1} = \frac{1 + \alpha T_2}{1 + 30\alpha} = \frac{11}{10} = 1.1$.$1 + \alpha T_2 = 1.1(1 + 30\alpha) = 1.1 + 33\alpha$.$\alpha T_2 = 0.1 + 33\alpha$.$T_2 = \frac{0.1}{\alpha} + 33$.Substituting $\alpha = 0.002$: $T_2 = \frac{0.1}{0.002} + 33 = 50 + 33 = 83^{\circ}C$.

The resistance of a wire at $30^{\circ}C$ is $10\, \Omega$. To increase the resistance of the wire by $10\%$,the temperature should be .......... $^{\circ}C$. The temperature coefficient of resistance of the material of the wire is $0.002\, ^{\circ}C^{-1}$.

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