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The resistance of a thin silver wire is ( $1.0 Ω a t 20 ° C$ ). The wire is placed in a liquid bath and its resistance rises to $1.2 Ω$ . What is the temperature of the bath? $α$ for silver is $3.8 \times 10^{- 3} / {}^{\circ}C$ .

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$T = 72.6 ° C$

$T = 89.6 ° C$

$T = 49.6 ° C$

$T = 116.6 ° C$

answer is C.

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Detailed Solution

$R (T) = R_{0} [1 + α (T - T_{0})]$

Here, $R (T) = 1.2 Ω, R_{0} = 1.0 Ω, α = 3.8 \times 10^{- 3} / {}^{\circ}C$ and $T_{0} = 20 ° C$

Substituting the values, we have $1.2 = 1.0 [1 + 3.8 \times 10^{- 3} (T - 20)]$

$3.8 \times 10^{- 3} (T - 20) = 0.2$

Solving this, we get

$T = 72.6 ° C$

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