A solid cylinder of denisty `rho_(0)`, cross-section area A and length `l` floats in a liquid `rho(gtrho_(0)` with its axis vertical, . If it is slightly displaced downward and released, the time period will be :

Options

A. `2pi sqrt((l)/(g))`
B. `2pi sqrt((rho_(0)l)/(rho g))`
C. `2pi sqrt((rho l)/(rho_(0)g))`
D. `2pi sqrt((l)/(2g))`

Answer

To solve the problem of finding the time period of a solid cylinder floating in a liquid, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Parameters:**
- Density of the cylinder: \( \rho_0 \)
- Density of the liquid: \( \rho \) (where \( \rho > \rho_0 \))
- Cross-sectional area of the cylinder: \( A \)
- Length of the cylinder: \( l \)

2. **Establish Equilibrium Condition:**
- When the cylinder is floating in equilibrium, the weight of the cylinder is equal to the buoyant force acting on it.
- Weight of the cylinder:
\[ W = \rho_0 \cdot g \cdot V = \rho_0 \cdot g \cdot (A \cdot l) \]
- Buoyant force:
\[ F_b = \rho \cdot g \cdot V_{displaced} = \rho \cdot g \cdot (A \cdot h) \]
- Setting these equal gives:
\[ \rho_0 \cdot g \cdot (A \cdot l) = \rho \cdot g \cdot (A \cdot h) \]
- Canceling \( g \) and \( A \) from both sides:
\[ \rho_0 \cdot l = \rho \cdot h \]
- Rearranging gives:
\[ h = \frac{\rho_0}{\rho} \cdot l \]

3. **Displace the Cylinder:**
- Let the cylinder be displaced downward by a small distance \( y \). The new submerged depth becomes \( h + y \).

4. **Calculate the Restoring Force:**
- The new buoyant force when displaced:
\[ F_b' = \rho \cdot g \cdot (A \cdot (h + y)) = \rho \cdot g \cdot (A \cdot h + A \cdot y) \]
- The change in buoyant force (restoring force) when displaced downward:
\[ F_{restoring} = F_b' - W = \rho \cdot g \cdot (A \cdot (h + y)) - \rho_0 \cdot g \cdot (A \cdot l) \]
- Substituting \( h \) from the equilibrium condition:
\[ F_{restoring} = \rho \cdot g \cdot (A \cdot \left(\frac{\rho_0}{\rho} \cdot l + y\right)) - \rho_0 \cdot g \cdot (A \cdot l) \]
- Simplifying gives:
\[ F_{restoring} = \rho \cdot g \cdot A \cdot y \]

5. **Use Newton's Second Law:**
- According to Newton's second law:
\[ F_{restoring} = m \cdot a \]
- The mass \( m \) of the cylinder is \( \rho_0 \cdot A \cdot l \):
\[ \rho \cdot g \cdot A \cdot y = \rho_0 \cdot A \cdot l \cdot a \]
- The acceleration \( a \) can be expressed as \( \frac{d^2y}{dt^2} \):
\[ \rho \cdot g \cdot y = \rho_0 \cdot l \cdot \frac{d^2y}{dt^2} \]

6. **Formulate the Differential Equation:**
- Rearranging gives:
\[ \frac{d^2y}{dt^2} + \frac{\rho g}{\rho_0 l} y = 0 \]
- This is a simple harmonic motion equation with angular frequency \( \omega^2 = \frac{\rho g}{\rho_0 l} \).

7. **Find the Time Period:**
- The time period \( T \) is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{\rho_0 l}{\rho g}} \]

### Final Answer:
The time period \( T \) of the oscillating cylinder is:
\[ T = 2\pi \sqrt{\frac{\rho_0 l}{\rho g}} \]