Statement I: A circular metal hop is suspended on the edge by a hook. the hoop can oscillate side to side in the plane of the hoop, or it can oscillate back and forth in a direction perpendicular to the plane of the hoop. The time period of oscillation would be more when oscillation are carried out in the plane of hoop.
Statement II: Time period of physical pendulum is more if moment of inertial of the rigid body about corresponding axis passing through the pivoted point is more.

Options

A. (a)Statement I is true statement II is true, Statement II is a correct explanation for Statement I.
B. (b)Statement I is true statement II is true, Statement II is NOT a correct explanation for Statement I.
C. (c)Statement I is true, Statement II is false
D. (d)Statement I is false, Statement II is true

Explanation

To solve the problem, we need to analyze both statements regarding the oscillation of a circular metal hoop and the relationship between the moment of inertia and the time period of oscillation.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- A circular metal hoop is suspended at its edge by a hook.
- The hoop can oscillate in two different ways:
- Side to side in the plane of the hoop.
- Back and forth in a direction perpendicular to the plane of the hoop.

2. **Time Period of Oscillation**:
- The time period \( T \) of a physical pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{I}{mgd}} \]
where:
- \( I \) is the moment of inertia about the pivot point,
- \( m \) is the mass of the hoop,
- \( g \) is the acceleration due to gravity,
- \( d \) is the distance from the pivot to the center of mass.

3. **Calculating Moment of Inertia**:
- For a circular hoop of mass \( m \) and radius \( R \):
- When oscillating in the plane of the hoop (let's denote this as \( I_1 \)), the moment of inertia about the pivot point (edge of the hoop) can be calculated using the parallel axis theorem:
\[ I_1 = I_{CM} + md^2 = \frac{1}{2}mR^2 + mR^2 = \frac{3}{2}mR^2 \]
- When oscillating perpendicular to the plane of the hoop (let's denote this as \( I_2 \)):
\[ I_2 = I_{CM} + md^2 = \frac{1}{2}mR^2 + mR^2 = 2mR^2 \]

4. **Comparing Time Periods**:
- The time period for oscillation in the plane of the hoop (using \( I_1 \)):
\[ T_1 = 2\pi \sqrt{\frac{I_1}{mgd}} = 2\pi \sqrt{\frac{\frac{3}{2}mR^2}{mgR}} = 2\pi \sqrt{\frac{3R}{2g}} \]
- The time period for oscillation perpendicular to the plane of the hoop (using \( I_2 \)):
\[ T_2 = 2\pi \sqrt{\frac{I_2}{mgd}} = 2\pi \sqrt{\frac{2mR^2}{mgR}} = 2\pi \sqrt{\frac{2R}{g}} \]

5. **Conclusion**:
- Since \( T_1 \) involves \( \sqrt{\frac{3R}{2g}} \) and \( T_2 \) involves \( \sqrt{\frac{2R}{g}} \), we can see that \( T_1 > T_2 \).
- Therefore, the time period of oscillation is greater when the oscillation is carried out in the plane of the hoop.

### Evaluation of Statements:
- **Statement I**: True. The time period of oscillation is more when oscillations are carried out in the plane of the hoop.
- **Statement II**: True. The time period of a physical pendulum is more if the moment of inertia about the corresponding axis passing through the pivot point is greater.

Thus, both statements are correct.