Earth may be assumed to be an axially symmetric freely rotating rigid body. The ratio of the principal moments of inertia about the axis of symmetry and an axis perpendicular to it is 33:32. If T is the time taken by earth to make one rotation around its axis of symmetry, then the time period of precession is closest to

Correct option is C

Given:
1. Earth is an axially symmetric, freely rotating rigid body.
2. The ratio of the principal moments of inertia about the axis of symmetry $$I_z$$​ and an axis perpendicular to it $$I_x = I_y$$​ is given as $$I_z : I_x = 33 : 32$$​ .
3.  T₀ is the time taken by Earth to make one rotation around its axis of symmetry.
The time period of precession  T  of a symmetric rigid body is given by:
​$$\\ T = \frac{2 \pi I_x}{(I_z - I_x) \omega} \\$$​​
where:
I_x  is the moment of inertia about an axis perpendicular to the axis of symmetry.
 I_z  is the moment of inertia about the axis of symmetry.
$$\ \omega \$$​is the angular velocity of Earth's rotation around its axis of symmetry.
Solution:
Expressing Moments of Inertia Using the Given Ratio:
- We are given $$\ I_z : I_x = 33 : 32$$​​
​$$\text{Let } I_x = 32k \text{ and } I_z = 33k \text{ for some constant } k.$$

$$\text{2. Calculate } I_z - I_x:$$

$$I_z - I_x = 33k - 32k = k$$

$$\text{3. Angular Velocity } \omega:$$ $$\text{- The angular velocity } \omega \text{ of Earth's rotation is related to the rotation period } T_0 \text{ by:}$$ $$\omega = \frac{2 \pi}{T_0}$$

$$\text{4. Substitute into the Precession Period Formula:}$$ $$\text{Substitute } I_x = 32k, \, I_z - I_x = k, \text{ and } \omega = \frac{2 \pi}{T_0} \text{ into the formula:}$$ $$T = \frac{2 \pi \cdot 32k}{k \cdot \frac{2 \pi}{T_0}}$$

$$T = \frac{2 \pi \cdot 32k \cdot T_0}{k \cdot 2 \pi} = 32 T_0$$

$$\text{Final Answer:}$$ $$\text{The time period of precession is closest to: } 32 \, T_0$$​​​​