The distance of point of a compound pendulum form its centre of gravity is `l`, the time period of oscillation relative to this point `T`. If `g = pi^(2)`, the relation between `l` and `T` will be :-

Options

A. `l^(2)-[T^(2)/4]l+k^(2)=0`
B. `l^(2)+[T^(2)/4]l+k^(2)=0`
C. `l^(2)-[T^(2)/4]l-k^(2)=0`
D. `l^(2)+[T^(2)/4]l-k^(2)=0`

Explanation

To find the relation between the distance \( l \) of a point of a compound pendulum from its center of gravity and the time period of oscillation \( T \), we can follow these steps:

### Step-by-step Solution:

1. **Understanding the Problem**:
We have a compound pendulum with a distance \( l \) from its center of gravity and we need to relate this distance to the time period \( T \) of oscillation. We are given that \( g = \pi^2 \).

2. **Torque Calculation**:
The torque \( \tau \) acting on the pivot point due to the weight of the pendulum can be expressed as:
\[ \tau = mg \sin \theta \cdot l \]
where \( mg \) is the weight of the pendulum and \( \theta \) is the angle of displacement.

3. **Moment of Inertia**:
The moment of inertia \( I \) of the pendulum about the pivot point can be expressed as:
\[ I = mk^2 + ml^2 \]
where \( k \) is the distance from the pivot to the center of mass.

4. **Angular Acceleration**:
The angular acceleration \( \alpha \) can be related to torque by:
\[ \tau = I \alpha \]
Substituting the expressions for torque and moment of inertia, we have:
\[ mg \sin \theta \cdot l = (mk^2 + ml^2) \alpha \]

5. **Simplifying the Equation**:
Dividing through by \( m \) (assuming \( m \neq 0 \)):
\[ g \sin \theta \cdot l = (k^2 + l^2) \alpha \]
Rearranging gives:
\[ \alpha = \frac{g \sin \theta \cdot l}{k^2 + l^2} \]

6. **Small Angle Approximation**:
For small angles, \( \sin \theta \approx \theta \):
\[ \alpha = \frac{g \theta \cdot l}{k^2 + l^2} \]

7. **Relating Angular Acceleration to Angular Frequency**:
We know that \( \alpha = \frac{d^2 \theta}{dt^2} = \omega^2 \theta \). Thus:
\[ \omega^2 = \frac{g l}{k^2 + l^2} \]

8. **Finding the Time Period**:
The relationship between angular frequency \( \omega \) and time period \( T \) is given by:
\[ \omega = \frac{2\pi}{T} \]
Therefore:
\[ \frac{2\pi}{T} = \sqrt{\frac{g l}{k^2 + l^2}} \]

9. **Squaring Both Sides**:
Squaring both sides gives:
\[ \left(\frac{2\pi}{T}\right)^2 = \frac{g l}{k^2 + l^2} \]

10. **Rearranging for \( T^2 \)**:
Rearranging leads to:
\[ T^2 = \frac{4\pi^2 (k^2 + l^2)}{g l} \]

11. **Substituting \( g = \pi^2 \)**:
Substituting \( g = \pi^2 \):
\[ T^2 = \frac{4\pi^2 (k^2 + l^2)}{\pi^2 l} \]
Simplifying gives:
\[ T^2 = \frac{4(k^2 + l^2)}{l} \]

12. **Final Relation**:
Rearranging leads to the final quadratic equation:
\[ l^2 - \frac{T^2}{4} l + k^2 = 0 \]

### Conclusion:
The relation between \( l \) and \( T \) is given by:
\[ l^2 - \frac{T^2}{4} l + k^2 = 0 \]