To solve this problem, we need to analyze the relation between the frequency of oscillations of the magnets, their moments of inertia, and their magnetic moments.

The frequency of oscillation of a magnet in a uniform magnetic field is given by:

$$ \nu = \frac{1}{2\pi} \sqrt{\frac{\mathrm{M} \mathrm{B}}{\mathrm{I}}} $$

Where:

$$\nu$$ is the frequency of oscillation
$$\mathrm{M}$$ is the magnetic moment of the magnet
$$\mathrm{B}$$ is the magnetic field strength
$$\mathrm{I}$$ is the moment of inertia of the magnet

For magnet 'P' and magnet 'Q', we can write:

$$ \nu_P = \frac{1}{2\pi} \sqrt{\frac{\mathrm{M_P} \mathrm{B}}{\mathrm{I_P}}} $$

$$ \nu_Q = \frac{1}{2\pi} \sqrt{\frac{\mathrm{M_Q} \mathrm{B}}{\mathrm{I_Q}}} $$

We are given that:

$$\nu_P = 2\nu_Q$$
$$\mathrm{I_P} = 2\mathrm{I_Q}$$

Substituting the given conditions into the frequency equations:

$$ \nu_P = 2 \nu_Q = 2 \left( \frac{1}{2\pi} \sqrt{\frac{\mathrm{M_Q} \mathrm{B}}{\mathrm{I_Q}}} \right) = \frac{1}{2\pi} \sqrt{4\frac{\mathrm{M_Q} \mathrm{B}}{\mathrm{I_Q}}} $$

Equating this to the expression for $$\nu_P$$, we get:

$$ \frac{1}{2\pi} \sqrt{\frac{\mathrm{M_P} \mathrm{B}}{\mathrm{I_P}}} = \frac{1}{2\pi} \sqrt{4\frac{\mathrm{M_Q} \mathrm{B}}{\mathrm{I_Q}}} $$

Squaring both sides:

$$ \frac{\mathrm{M_P} \mathrm{B}}{\mathrm{I_P}} = 4 \frac{\mathrm{M_Q} \mathrm{B}}{\mathrm{I_Q}} $$

Given that $$\mathrm{I_P} = 2\mathrm{I_Q}$$, substitute this into the equation:

$$ \frac{\mathrm{M_P} \mathrm{B}}{2\mathrm{I_Q}} = 4 \frac{\mathrm{M_Q} \mathrm{B}}{\mathrm{I_Q}} $$

Simplify:

$$ \frac{\mathrm{M_P}}{2} = 4 \mathrm{M_Q} $$

So:

$$ \mathrm{M_P} = 8 \mathrm{M_Q} $$

Therefore, the correct option is:

Option C: $$\mathrm{M}_{\mathrm{P}}=8 \mathrm{M_Q}$$

Magnetism And Matter

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