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((lrho)/((rho-d)g))`
B. `2pisqrt((ld)/(rhog))`
C. `2pisqrt((lrho)/(dg))`
D. `2pisqrt((ld)/((rho-d)g))`

Explanation

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

### Step 1: Identify the mass of the cube
The mass \( m \) of the wooden cube can be calculated using the formula:
\[ m = \text{Volume} \times \text{Density} \]
The volume \( V \) of the cube with side length \( l \) is:
\[ V = l^3 \]
Thus, the mass \( m \) of the cube is:
\[ m = l^3 \cdot d \]
where \( d \) is the density of the wood.

### Step 2: Determine the area of the cube's base
The area \( A \) of the base of the cube is given by:
\[ A = l^2 \]

### Step 3: Calculate the restoring force
When the cube is pushed down slightly, the weight of the displaced liquid creates a restoring force. The volume of the submerged part of the cube when pushed down by a small distance \( x \) is:
\[ \text{Volume submerged} = A \cdot x = l^2 \cdot x \]
The weight of the displaced liquid is:
\[ \text{Weight of displaced liquid} = \text{Volume submerged} \times \text{Density of liquid} \times g = l^2 \cdot x \cdot \rho \cdot g \]
The restoring force \( F \) acting on the cube is equal to the weight of the displaced liquid:
\[ F = l^2 \cdot x \cdot \rho \cdot g \]

### Step 4: Relate the restoring force to acceleration
According to Hooke's law, the restoring force is also related to the acceleration \( a \) of the cube:
\[ F = -k x \]
where \( k \) is the spring constant. The acceleration \( a \) can be expressed as:
\[ a = \frac{F}{m} \]
Substituting the expression for \( F \):
\[ a = \frac{l^2 \cdot x \cdot \rho \cdot g}{l^3 \cdot d} = \frac{l^2 \cdot \rho \cdot g}{l^3 \cdot d} \cdot x = \frac{\rho \cdot g}{l \cdot d} \cdot x \]

### Step 5: Find the period of simple harmonic motion
The period \( T \) of simple harmonic motion is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
From our previous steps, we have:
\[ k = \frac{\rho \cdot g}{l \cdot d} \]
Thus, substituting for \( m \) and \( k \):
\[ T = 2\pi \sqrt{\frac{l^3 \cdot d}{l^2 \cdot \rho \cdot g}} = 2\pi \sqrt{\frac{l \cdot d}{\rho \cdot g}} \]

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