MHT CET · Physics · Magnetic Properties of Matter
Above the curie temperature the susceptibility of a ferromagnetic substance varies
- A directly as the absolute temperature.
- B inversely as the absolute temperature.
- C inversely as the square root of absolute temperature.
- D directly as the square root of absolute temperature.
Answer & Solution
Correct Answer
(B) inversely as the absolute temperature.
Step-by-step Solution
Detailed explanation
\(\chi \propto \frac{1}{\mathrm{~T}}\) (Theory question)
15. Gain in kinetic energy \(=\) loss in potential energy
\(
\begin{aligned}
& \frac{1}{2} \mathrm{~m}\left(1+\frac{\mathrm{k}^{2}}{\mathrm{R}^{2}}\right) \mathrm{V}^{2}=\mathrm{mgh} \\
& \frac{1}{2}\left(1+\frac{1}{2}\right) \mathrm{V}^{2}=\mathrm{gh} \\
\therefore & \frac{3}{4} \mathrm{~V}^{2}=\mathrm{gh} \\
\therefore & \mathrm{h}=\frac{3 \mathrm{~V}^{2}}{4 \mathrm{~g}}
\end{aligned}
\)
15. Gain in kinetic energy \(=\) loss in potential energy
\(
\begin{aligned}
& \frac{1}{2} \mathrm{~m}\left(1+\frac{\mathrm{k}^{2}}{\mathrm{R}^{2}}\right) \mathrm{V}^{2}=\mathrm{mgh} \\
& \frac{1}{2}\left(1+\frac{1}{2}\right) \mathrm{V}^{2}=\mathrm{gh} \\
\therefore & \frac{3}{4} \mathrm{~V}^{2}=\mathrm{gh} \\
\therefore & \mathrm{h}=\frac{3 \mathrm{~V}^{2}}{4 \mathrm{~g}}
\end{aligned}
\)
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