The distance between the point of suspension and the centre of gravity of a compound pendulum is `l` and the radius of gyration about the horizontal axis through the centre of gravity is `k`, then its time period will be

Options

A. `2pisqrt((l+k)/(g))`
B. `2pisqrt((l^(2)+k^(2))/(lg))`
C. `2pisqrt((l+k^(2))/(g))`
D. `2pisqrt((2k)/(lg))`

Explanation

To find the time period of a compound pendulum, we can use the following steps:

### Step 1: Understand the Components
We have a compound pendulum with:
- Distance from the point of suspension to the center of gravity = \( l \)
- Radius of gyration about the horizontal axis through the center of gravity = \( k \)

### Step 2: Use the Formula for Time Period
The time period \( T \) of a compound pendulum can be expressed using the formula:
\[ T = 2\pi \sqrt{\frac{I}{\tau}} \]
where \( I \) is the moment of inertia about the pivot point and \( \tau \) is the torque due to gravity.

### Step 3: Calculate the Moment of Inertia
According to the parallel axis theorem, the moment of inertia \( I \) about the point of suspension can be calculated as:
\[ I = m k^2 + m l^2 \]
where \( m \) is the mass of the pendulum.

### Step 4: Calculate the Torque
The torque \( \tau \) about the pivot due to the weight of the pendulum is given by:
\[ \tau = mg \cdot l \]
where \( g \) is the acceleration due to gravity.

### Step 5: Substitute into the Time Period Formula
Now, substituting \( I \) and \( \tau \) into the time period formula:
\[ T = 2\pi \sqrt{\frac{m k^2 + m l^2}{mg \cdot l}} \]

### Step 6: Simplify the Expression
We can simplify this expression by canceling \( m \):
\[ T = 2\pi \sqrt{\frac{k^2 + l^2}{g \cdot l}} \]

### Final Result
Thus, the time period \( T \) of the compound pendulum is:
\[ T = 2\pi \sqrt{\frac{l^2 + k^2}{g \cdot l}} \]