Resistance of the wire is measured as 2Ω and 3Ω at 10∘C and 30∘C respectively. The resistance of the wire at 0 C o is

R=R0(1+αΔT)3=R0(1+α(30−0))2=R0(1+α(10−0))32=1+30α1+10αα=130=0.033 C −1 oTherefore3=R01+130(30−0)⇒R0=32Ω

Resistance of the wire is measured as 2Ω and 3Ω at 10∘C and 30∘C respectively. The resistance of the wire at 0 C o is

R=R0(1+αΔT)3=R0(1+α(30−0))2=R0(1+α(10−0))32=1+30α1+10αα=130=0.033 C −1 oTherefore3=R01+130(30−0)⇒R0=32Ω

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Resistance of the wire is measured as $2 Ω and 3 Ω at 10^{\circ} C and 30^{\circ} C$ respectively. The resistance of the wire at $0 {}^{o}C$ is

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$2 Ω$

$1.5 Ω$

$1 Ω$

$0.33 Ω$

answer is B.

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Detailed Solution

$\begin{array}{l} R = R_{0} (1 + αΔT) \\ 3 = R_{0} (1 + α (30 - 0)) \\ 2 = R_{0} (1 + α (10 - 0)) \\ \frac{3}{2} = \frac{1 + 30 α}{1 + 10 α} \\ α = \frac{1}{30} = 0 .033 {}^{o}C^{- 1} \end{array}$

Therefore

$\begin{array}{l} 3 = R_{0} (1 + \frac{1}{30} (30 - 0)) \end{array}$

$\Rightarrow R_{0} = \frac{3}{2} Ω$

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Resistance of the wire is measured as 2Ω and 3Ω at 10∘C and 30∘C respectively. The resistance of the wire at 0 C o is

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Resistance of the wire is measured as $2 Ω and 3 Ω at 10^{\circ} C and 30^{\circ} C$ respectively. The resistance of the wire at $0 {}^{o}C$ is