KCET · Physics · Magnetic Properties of Matter
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 \(1 / 3\)
- B \(3\)
- C \(7 / 5\)
- D \(5 / 7\)
Answer & Solution
Correct Answer
(B) \(3\)
Step-by-step Solution
Detailed explanation
We know that, magnetic susceptibility,
\(\chi=\frac{C}{T-T_C}\)
\(\therefore \chi \propto \frac{1}{T-T_C}\)
\(\Rightarrow \frac{\chi_{\text {cobalt }}}{\chi_{\text {iron }}} =\frac{T-\left(T_C\right)_{\text {iron }}}{T-\left(T_C\right)_{\text {cobalt }}}\)
\(=\frac{1600-1000}{1600-1400}=\frac{600}{200}=3\)
\(\chi=\frac{C}{T-T_C}\)
\(\therefore \chi \propto \frac{1}{T-T_C}\)
\(\Rightarrow \frac{\chi_{\text {cobalt }}}{\chi_{\text {iron }}} =\frac{T-\left(T_C\right)_{\text {iron }}}{T-\left(T_C\right)_{\text {cobalt }}}\)
\(=\frac{1600-1000}{1600-1400}=\frac{600}{200}=3\)
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