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The resistance of a platinum wire is $2.4$ ohms at $0^\circ C$ and $3.4$ ohms at$100^\circ C$. If the resistance at $t^\circ C$ is 4 ohms then the temperature $t$ is equal to
A) $160^\circ C$
B) $140^\circ C$
C) $180^\circ C$
D) $120^\circ C$

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Last updated date: 13th Jun 2024
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Answer
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Hint The resistance of a wire rises when its temperature rises. Use the resistance of the wire given at different points to determine the temperature coefficient of resistance. Then use the temperature coefficient of resistance and calculate the temperature rise required for resistance of 4 ohms.
$\Rightarrow \Delta R = \alpha {R_0}\Delta T$ where $\Delta R$ is the change in resistance of the wire, $\alpha $ is the temperature coefficient of resistance, ${R_0}$ is the reference temperature, and $\Delta T$ is the change in temperature

Complete step by step answer
The rise in resistance of a cable is given by
$\Rightarrow \Delta R = \alpha {R_0}\Delta T$
Since we don’t know the temperature coefficient of resistance $\alpha $, let us calculate it using the data given in the question.
We know that the resistance of the platinum wire is $2.4$ ohms at $0^\circ C$ and $3.4$ ohms at $100^\circ C$. Using $0^\circ C$ as our reference temperature, we can use the formula $\Delta R = \alpha {R_0}\Delta T$ and write
$\Rightarrow 3.4 - 2.4 = \alpha \times 2.4 \times 100$
So the value of $\alpha $ comes out as:
$\Rightarrow \alpha = 0.00416\,/^\circ C$
Using the value of the temperature coefficient of resistance that we just calculated, let us now determine the temperature at which the resistance is 4 ohms again using $0^\circ C$ as our reference temperature.
So substituting $\Delta R = 4 - 2.4$ and $\Delta T = t - 0$ in $\Delta R = \alpha {R_0}\Delta T$, we get
$\Rightarrow 4 - 2.4 = 0.00416 \times 2.4 \times t$
Solving for $t$, we get
$\Rightarrow t = 160^\circ C$
Hence the temperature of the wire must be $160^\circ C$ for its resistance to be 4 ohms.

Note
To calculate the change in resistance due to a temperature change, we must have the temperature coefficient of resistance which hasn’t been given directly in the question. But it can be derived from the given data of resistance values at two given temperature values which we should realize to use.