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**Hint:**To solve this problem, use the relationship between temperature and resistance of conductor which is given as ${R}_{t}= {R}_{0}(1 + \alpha \Delta t)$. Substitute the values in this formula for both the resistances and add them up as both the resistances are connected in series. Solve the obtained equation and find the effective temperature of the combination by comparing with the general formula mentioned above.

**Formula used:**

${R}_{t}= {R}_{0}(1 + \alpha \Delta t)$

**Complete answer:**

Given: ${R}_{1}= 600\Omega$

${R}_{2}= 300\Omega$

${\alpha}_{1}= 0.001 {K}^{-1}$

${\alpha}_{2}= 0.004 {K}^{-1}$

The relationship between temperature and resistance of conductor is given by,

${R}_{t}= {R}_{0}(1 + \alpha \Delta t)$ …(1)

Where, ${R}_{t}$ is the resistance at temperature t

${R}_{0}$ is the resistance at temperature usually 20°C

$\alpha$ is the temperature coefficient of resistance

$\Delta t$ is the difference between operating and reference temperature

Resistances are connected in series. So, their equivalent resistance is given by,

${R}_{eq}= {R}_{t1}+ {R}_{t2}$ …(2)

Using equation. (1), equation. (2) can be written as,

${R}_{eq}= {R}_{1}(1 + {\alpha}_{1} \Delta t) + {R}_{2}(1 + {\alpha}_{2} \Delta t)$

Substituting values in above equation we get,

${R}_{eq}= 600 (1+ 0.001 \Delta t)+ 300(1+ 0.004 \Delta t)$

$\Rightarrow {R}_{eq}= 600 + 0.6 \Delta t + 300 + 1.2 \Delta t$

$\Rightarrow {R}_{eq}= 900+ 1.8 \Delta t$

$\Rightarrow {R}_{eq}= 900 (1 + 0.002 \Delta t)$ …(3)

Comparing equation. (3) with equation. (1) we get,

$\alpha= 0.002 {K}^{-1}$

$\Rightarrow \alpha= \dfrac {1}{500} {degrees}^{-1}$

Hence, the effective temperature of the combination is $\dfrac {1}{500} {degrees}^{-1}$.

**So, the correct answer is option C i.e. $\dfrac {1}{500}{degrees}^{-1}$.**

**Note:**

If the temperature coefficient of resistance is positive for a material then it means that the resistance increases with increase in temperature. Whereas, if the temperature coefficient of resistance is negative for a material then it means that the resistance decreases with increase in temperature. Pure metals have positive temperature coefficient of resistance whereas semiconductors and insulators have negative temperature coefficient of resistance.

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