A thin uniform rod of length l is pivoted at its upper end. It is free to swing in a vertical plane. Its time period for oscillation of small amplitude is

Options

A. `2pisqrt((l)/(g))`
B. `2pisqrt((2l)/(3g))`
C. `2pisqrt((3l)/(2g))`
D. `2pisqrt((l)/(3g))`

Explanation

To find the time period of a thin uniform rod of length \( l \) pivoted at its upper end and swinging in a vertical plane, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Type of Pendulum**:
The rod acts as a physical pendulum since it is pivoted at one end and can swing back and forth.

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

3. **Calculate the Moment of Inertia**:
For a thin uniform rod pivoted at one end, the moment of inertia \( I \) is given by:
\[ I = \frac{1}{3} m l^2 \]

4. **Determine the Distance to the Center of Mass**:
The center of mass of a uniform rod is located at its midpoint, which is at a distance of:
\[ h = \frac{l}{2} \]

5. **Substitute Values into the Time Period Formula**:
Now substituting \( I \) and \( h \) into the time period formula:
\[ T = 2\pi \sqrt{\frac{\frac{1}{3} m l^2}{mg \cdot \frac{l}{2}}} \]

6. **Simplify the Expression**:
Cancel out \( m \) from the numerator and denominator:
\[ T = 2\pi \sqrt{\frac{\frac{1}{3} l^2}{g \cdot \frac{l}{2}}} \]
This simplifies to:
\[ T = 2\pi \sqrt{\frac{2l}{3g}} \]

7. **Final Result**:
Thus, the time period for small oscillations of the rod is:
\[ T = 2\pi \sqrt{\frac{2l}{3g}} \]

### Conclusion:
The correct answer for the time period of oscillation of the rod is:
\[ T = 2\pi \sqrt{\frac{2l}{3g}} \]