Q171 / 180 · KCET 2023 · Physics
The Curie temperature of cobalt and iron are $1400 \mathrm{~K}$ and $1000 \mathrm{~K}$ respectively. At $T=1600 \mathrm{~K}$, the ratio of magnetic susceptibility of cobalt to that of iron is
- A$1 / 3$
- Correct answerB$3$
- C$7 / 5$
- D$5 / 7$
Solution
We know that, magnetic susceptibility,
$\begin{aligned} & \chi=\frac{C}{T-T_C} \\ \therefore \quad & \chi \propto \frac{1}{T-T_C}\end{aligned}$
$\begin{aligned} \Rightarrow \frac{\chi_{\text {cobalt }}}{\chi_{\text {iron }}} & =\frac{T-\left(T_C\right)_{\text {iron }}}{T-\left(T_C\right)_{\text {cobalt }}} \\ & =\frac{1600-1000}{1600-1400}=\frac{600}{200}=3\end{aligned}$
Hence, the correct answer is option: 'B'.