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When the temperature of a heating element changes, its resistance also changes and so does the temperature coefficient. Suppose the temperature varies linearly with time and is given by $T (° C) = (20 + 10 t)$ where $t$ is time in second. The temperature coefficient of the material is $0.0065 {(° C)}^{- 1}$ at $0 ° C$ . If the initial resistance of the heating element is $2 r$ , then express the resistance as a function of time.

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$R (t) = (2 + 0.115 t) Ω$

$R (t) = (5 + 0.105 t) Ω$

$R (t) = (8 + 0.355 t) Ω$

$R (t) = (7 + 0.915 t) Ω$

answer is A.

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Detailed Solution

Resistance at temperature $T$ is given by

$R_{T} = R_{0} (1 + α_{0} T) = R (t)$

From the given data,

$R_{0} = R_{1} [1 + α_{1} (0 - 20 °)] R_{0} = 2 (1 - 20 \times 0.00575) R_{0} = 1.77 Ω$

Since, $T = 20 + 10 t$

$R (t) = 1.77 [1 + 0.0065 (20 + 10 t)] R (t) = (2 + 0.115 t) Ω$

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