The string of a simple pendulum replaced by a uniform rod of length `L` and mass `M` while the bob has a mass `m`. It is allowed to make small oscillation. Its time period is

Options

A. `2pisqrt(((2M)/(3m))(L)/(g))`
B. `2pisqrt((2(M+3m)L)/(3(M+2m)g))`
C. `2pisqrt(((M+m)/(M+3m))(L)/(g))`
D. `2pisqrt(((2m+M)/(3(M+2m)))(L)/(g))`

Explanation

To find the time period of a simple pendulum where the string is replaced by a uniform rod of length \( L \) and mass \( M \), with a bob of mass \( m \), we can follow these steps:

### Step-by-Step Solution

1. **Identify the Formula for Time Period**:
The time period \( T \) of a physical pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{I_{\text{net}}}{M_{\text{total}} \cdot g \cdot d}} \]
where:
- \( I_{\text{net}} \) is the moment of inertia about the pivot point,
- \( M_{\text{total}} \) is the total mass,
- \( g \) is the acceleration due to gravity,
- \( d \) is the distance from the pivot to the center of mass.

2. **Calculate the Moment of Inertia**:
The moment of inertia for the rod about one end is given by:
\[ I_{\text{rod}} = \frac{1}{3} M L^2 \]
The moment of inertia for the bob (considered as a point mass) at a distance \( L \) from the pivot is:
\[ I_{\text{bob}} = m L^2 \]
Therefore, the total moment of inertia \( I_{\text{net}} \) is:
\[ I_{\text{net}} = I_{\text{rod}} + I_{\text{bob}} = \frac{1}{3} M L^2 + m L^2 \]

3. **Calculate the Center of Mass**:
The center of mass \( d \) of the system (rod + bob) can be calculated using the formula:
\[ d = \frac{M \cdot \frac{L}{2} + m \cdot L}{M + m} \]
This simplifies to:
\[ d = \frac{ML/2 + mL}{M + m} = \frac{(M/2 + m)L}{M + m} \]

4. **Substituting Values into the Time Period Formula**:
Now we substitute \( I_{\text{net}} \), \( M_{\text{total}} \), \( g \), and \( d \) into the time period formula:
\[ T = 2\pi \sqrt{\frac{\frac{1}{3} M L^2 + m L^2}{(M + m)g \cdot \frac{(M/2 + m)L}{M + m}}} \]
Simplifying this expression will give us the final form of the time period.

5. **Final Expression**:
After simplification, we arrive at:
\[ T = 2\pi \sqrt{\frac{2(M + 3m)L}{3(M + 2m)g}} \]

### Final Answer
The time period \( T \) of the pendulum is:
\[ T = 2\pi \sqrt{\frac{2(M + 3m)L}{3(M + 2m)g}} \]