A rectangular block of mass m and area of cross-section A floats in a liquid of density `rho`. If it is givan a small vertical displacement from equilibrium, it undergoes oscillation with a time period T, then select the wrong alternative.

Options

A. `T^(2) alpha m`
B. `T^(2)alpha g`
C. `T^(2) alpha 1//A`
D. `T^(2)alpha 1/(rho)`

Answer

To solve the problem step by step, we will analyze the given information about the rectangular block floating in a liquid and derive the time period of oscillation.

### Step 1: Understand the equilibrium condition
When the block is floating in the liquid, the buoyant force acting on it is equal to its weight. According to Archimedes' principle, the buoyant force \( F_b \) is given by:
\[ F_b = \rho \cdot V \cdot g \]
where \( \rho \) is the density of the liquid, \( V \) is the volume of the liquid displaced, and \( g \) is the acceleration due to gravity.

### Step 2: Relate volume to area and length
The volume \( V \) of the block can be expressed in terms of its cross-sectional area \( A \) and the submerged length \( l \):
\[ V = A \cdot l \]
Thus, the buoyant force can be rewritten as:
\[ F_b = \rho \cdot (A \cdot l) \cdot g \]

### Step 3: Set up the equilibrium equation
At equilibrium, the weight of the block \( mg \) is balanced by the buoyant force:
\[ mg = \rho \cdot (A \cdot l) \cdot g \]
From this, we can simplify and solve for \( l \):
\[ l = \frac{mg}{\rho A g} = \frac{m}{\rho A} \]

### Step 4: Determine the time period of oscillation
For small vertical displacements, the time period \( T \) of oscillation can be derived from the formula:
\[ T = 2\pi \sqrt{\frac{l}{g}} \]
Substituting the expression for \( l \):
\[ T = 2\pi \sqrt{\frac{m}{\rho A g}} \]

### Step 5: Analyze the relationships
From the derived formula for \( T \):
- \( T^2 \) is directly proportional to \( m \) (mass of the block).
- \( T^2 \) is inversely proportional to \( A \) (area of cross-section).
- \( T^2 \) is inversely proportional to \( \rho \) (density of the liquid).
- \( T^2 \) is inversely proportional to \( g \) (acceleration due to gravity).

### Step 6: Identify the wrong alternative
Based on the relationships derived:
1. \( T^2 \) is directly proportional to \( m \) - **True**
2. \( T^2 \) is directly proportional to \( g \) - **False** (it is inversely proportional)
3. \( T^2 \) is inversely proportional to \( A \) - **True**
4. \( T^2 \) is inversely proportional to \( \rho \) - **True**

Thus, the wrong alternative is option 2.