Question 22725

Nov 2, 2016

$8 \Omega$

Explanation:

Resistance ${R}_{T}$ of a metallic wire is expressed in terms of its temperature $T$, stated in "^@C, as

R_T=R_"ref"[1+alpha(T-T_"ref")] ....(1)
where ${R}_{\text{ref}}$ is resistance at reference temperature ${T}_{\text{ref}}$, usually ${20}^{\circ} \text{C} .$ Sometimes ${0}^{\circ} \text{C}$ and
$\alpha$ is temperature coefficient of the conductor material.
Let us assume ${T}_{\text{ref}}$ be ${0}^{\circ} \text{C}$, from (1) we obtain
${R}_{0} = 5 \Omega$
To calculate $\alpha$ we use given resistance at ${100}^{\circ} \text{C}$. Again from (1)
7=5[1+alpha(100-0]#
$\implies \alpha = \frac{\frac{7}{5} - 1}{100}$
$\implies \alpha = \frac{2}{500}$
$\implies \alpha = 0.004$

To calculate resistance at ${150}^{\circ} \text{C}$, again from (1)
${R}_{150} = 5 \left[1 + 0.004 \left(150 - 0\right)\right]$
${R}_{150} = 8 \Omega$