A solid cube of side a and density `rho_(0)` floats on the surface of a liquid of density `rho`. If the cube is slightly pushed downward, then it oscillates simple harmonically with a period of

Options

A. (a)`2pisqrt((rho_(0))/(rho)(a)/(g))`
B. (b)`2pi sqrt((rho)/(rho_(0))(a)/(g))`
C. (c)`2pi sqrt ((a)/((1 - (rho)/(rho_(0)))g`
D. (d)`2pi sqrt ((a)/((1 + (rho)/(rho_(0)))g`

Explanation

To find the period of oscillation of a solid cube floating on the surface of a liquid when it is slightly pushed downward, we can follow these steps:

### Step 1: Understand the Forces Acting on the Cube
When the cube is floating, it displaces a volume of liquid equal to the weight of the cube. The buoyant force acting on the cube is equal to the weight of the liquid displaced.

### Step 2: Define the Variables
- Let the side of the cube be \( a \).
- Let the density of the cube be \( \rho_0 \).
- Let the density of the liquid be \( \rho \).
- Let \( L \) be the length of the cube that is initially submerged in the liquid.

### Step 3: Calculate the Buoyant Force
The buoyant force \( F_b \) acting on the cube when it is submerged to a depth \( L \) is given by:
\[ F_b = \rho \cdot a \cdot L \cdot g \]
where \( g \) is the acceleration due to gravity.

### Step 4: Set Up the Equilibrium Condition
At equilibrium, the buoyant force equals the weight of the cube:
\[ F_b = m \cdot g = \rho_0 \cdot a^3 \cdot g \]
From this, we can express the submerged length \( L \):
\[ \rho \cdot a \cdot L \cdot g = \rho_0 \cdot a^3 \cdot g \]
This simplifies to:
\[ \rho \cdot L = \rho_0 \cdot a^2 \]
Thus, we find:
\[ L = \frac{\rho_0 a^2}{\rho} \]

### Step 5: Analyze the Displacement
When the cube is pushed downward by a small displacement \( x \), the new submerged length becomes \( L + x \). The new buoyant force is:
\[ F_b' = \rho \cdot a \cdot (L + x) \cdot g \]
The net force acting on the cube when displaced is given by:
\[ F_{net} = F_b' - mg \]
Substituting the expressions we have:
\[ F_{net} = \rho \cdot a \cdot (L + x) \cdot g - \rho_0 \cdot a^3 \cdot g \]

### Step 6: Simplify the Net Force
Substituting \( L \):
\[ F_{net} = \rho \cdot a \cdot \left(\frac{\rho_0 a^2}{\rho} + x\right) g - \rho_0 \cdot a^3 \cdot g \]
This simplifies to:
\[ F_{net} = \rho_0 \cdot a^3 \cdot g + \rho \cdot a \cdot x \cdot g - \rho_0 \cdot a^3 \cdot g \]
Thus:
\[ F_{net} = \rho \cdot a \cdot x \cdot g \]

### Step 7: Relate Force to Acceleration
Using Newton's second law, we have:
\[ F_{net} = m \cdot a = \rho_0 \cdot a^3 \cdot \frac{d^2x}{dt^2} \]
Setting the two expressions for \( F_{net} \) equal gives:
\[ \rho \cdot a \cdot x \cdot g = \rho_0 \cdot a^3 \cdot \frac{d^2x}{dt^2} \]

### Step 8: Formulate the Equation of Motion
Rearranging gives:
\[ \frac{d^2x}{dt^2} + \frac{\rho g}{\rho_0 a} x = 0 \]
This is the equation of simple harmonic motion.

### Step 9: Identify the Angular Frequency
From the standard form of SHM:
\[ \omega^2 = \frac{\rho g}{\rho_0 a} \]
Thus, the angular frequency \( \omega \) is:
\[ \omega = \sqrt{\frac{\rho g}{\rho_0 a}} \]

### Step 10: Calculate the Time Period
The time period \( T \) of the oscillation is given by:
\[ T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{\rho_0 a}{\rho g}} \]

### Final Answer
The time period of the simple harmonic motion of the cube is:
\[ T = 2\pi \sqrt{\frac{\rho_0 a}{\rho g}} \]