A rod whose ends are `A` and `B` of length `25 cm` is hanged in vertical plane. When hanged from point `A` and point `B` the time periods calculated are `3`sec and `4`sec respectively. Given the moment of inertia of rod about axis perpendicular to the rod is in ratio `9:4` at points `A` and `B`. find the distance of the crntre of mass from point `A`.

Options

A. `9 cm`
B. `5 cm`
C. `25 cm`
D. `20 cm`

Answer

To solve the problem, we need to use the formula for the time period of a compound pendulum, which is given by:

\[ T = 2\pi \sqrt{\frac{I}{mgL}} \]

where:
- \( T \) is the time period,
- \( I \) is the moment of inertia about the pivot point,
- \( m \) is the mass of the rod,
- \( g \) is the acceleration due to gravity,
- \( L \) is the distance from the pivot point to the center of mass.

### Step-by-Step Solution

1. **Identify the Given Values:**
- Length of the rod, \( L_{rod} = 25 \, \text{cm} = 0.25 \, \text{m} \)
- Time period when hanged from point A, \( T_A = 3 \, \text{s} \)
- Time period when hanged from point B, \( T_B = 4 \, \text{s} \)
- Moment of inertia ratio \( \frac{I_A}{I_B} = \frac{9}{4} \)

2. **Express the Moments of Inertia:**
- Let \( I_A = 9x \) and \( I_B = 4x \).

3. **Set Up the Time Period Equations:**
- For point A:
\[ T_A = 2\pi \sqrt{\frac{I_A}{mgL_A}} \implies 3 = 2\pi \sqrt{\frac{9x}{mgL_A}} \]
- For point B:
\[ T_B = 2\pi \sqrt{\frac{I_B}{mgL_B}} \implies 4 = 2\pi \sqrt{\frac{4x}{mgL_B}} \]

4. **Square Both Equations to Eliminate the Square Root:**
- For point A:
\[ 9 = \frac{4\pi^2 \cdot 9x}{mgL_A} \implies mgL_A = 4\pi^2 \cdot 9x \implies L_A = \frac{36\pi^2 x}{mg} \]
- For point B:
\[ 16 = \frac{4\pi^2 \cdot 4x}{mgL_B} \implies mgL_B = 4\pi^2 \cdot 4x \implies L_B = \frac{16\pi^2 x}{mg} \]

5. **Relate \( L_A \) and \( L_B \) to the Length of the Rod:**
- Since the center of mass is at a distance \( d \) from point A, we have:
\[ L_A = d \quad \text{and} \quad L_B = 25 - d \]

6. **Substitute \( L_A \) and \( L_B \) into the Equations:**
- From point A:
\[ d = \frac{36\pi^2 x}{mg} \]
- From point B:
\[ 25 - d = \frac{16\pi^2 x}{mg} \]

7. **Combine the Equations:**
- Substitute \( d \) from the first equation into the second:
\[ 25 - \frac{36\pi^2 x}{mg} = \frac{16\pi^2 x}{mg} \]
- Rearranging gives:
\[ 25 = \frac{36\pi^2 x}{mg} + \frac{16\pi^2 x}{mg} = \frac{52\pi^2 x}{mg} \]
- Thus,
\[ mg = \frac{52\pi^2 x}{25} \]

8. **Substituting Back to Find \( d \):**
- Substitute \( mg \) back into the equation for \( d \):
\[ d = \frac{36\pi^2 x}{\frac{52\pi^2 x}{25}} = \frac{36 \cdot 25}{52} = \frac{900}{52} \approx 17.31 \, \text{cm} \]

### Final Answer:
The distance of the center of mass from point A is approximately \( 17.31 \, \text{cm} \).