A wooden cube (density of wood `'d'`) of side `'l'` flotes in a liquid of density `'rho'` with its upper and lower surfaces horizonta. If the cube is pushed slightly down and released, it performs simple harmonic motion of period `'T'`. Then, `'T'` is equal to :-

Options

A. `2pisqrt(1p/(p-d)g)`
B. `2pisqrt(1p/pg)`
C. `2pisqrt(1p/dg)`
D. `2pisqrt(1p/(p-d)g)`

Explanation

To find the time period \( T \) of the simple harmonic motion (SHM) of a wooden cube floating in a liquid, we can follow these steps:

### Step 1: Understand the problem
The wooden cube has a density \( d \) and side length \( l \). It floats in a liquid with density \( \rho \). When the cube is pushed down slightly and released, it performs SHM.

### Step 2: Identify the relevant formulas
The time period \( T \) of a simple harmonic motion can be expressed as:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
where \( m \) is the mass of the object and \( k \) is the effective spring constant.

### Step 3: Calculate the mass of the cube
The mass \( m \) of the wooden cube can be calculated using its volume and density:
\[ m = \text{Volume} \times \text{Density} = l^3 \cdot d \]

### Step 4: Determine the effective spring constant \( k \)
When the cube is displaced downwards, the buoyant force acting on it changes. The buoyant force \( F_b \) can be expressed as:
\[ F_b = \rho \cdot V \cdot g \]
where \( V \) is the submerged volume of the cube. For a cube of side \( l \), if it is submerged by a depth \( x \), the submerged volume \( V \) is:
\[ V = l^2 \cdot x \]
Thus, the buoyant force becomes:
\[ F_b = \rho \cdot (l^2 \cdot x) \cdot g \]

### Step 5: Find the effective spring constant \( k \)
The effective spring constant \( k \) can be derived from the relationship between the buoyant force and the displacement \( x \):
\[ F_b = kx \]
From the buoyant force equation:
\[ k = \frac{F_b}{x} = \frac{\rho \cdot l^2 \cdot g}{x} \]
This implies that the effective spring constant \( k \) is:
\[ k = \rho \cdot l^2 \cdot g \]

### Step 6: Substitute \( m \) and \( k \) into the time period formula
Now, substituting \( m \) and \( k \) into the time period formula:
\[ T = 2\pi \sqrt{\frac{l^3 \cdot d}{\rho \cdot l^2 \cdot g}} \]

### Step 7: Simplify the expression
This simplifies to:
\[ T = 2\pi \sqrt{\frac{l \cdot d}{\rho \cdot g}} \]

### Final Result
Thus, the time period \( T \) of the simple harmonic motion of the wooden cube is:
\[ T = 2\pi \sqrt{\frac{l \cdot d}{\rho \cdot g}} \]