A rectangular block of mass `m` and area of cross-section `A` floats in a liquid of density `rho`. If it is given a small vertical displacement from equilibrium, it undergoes oscillation with a time period `T`.
(i) `T prop sqrt(m)`
(ii) `T prop sqrt(rho)`
(iii) `T prop (1)/(sqrt(A))`
(iv) `T prop (1)/(sqrt(rho))`.

Options

A. (i),(ii),(iii)
B. (i),(iii),(iv)
C. (i),(iii)
D. (iii),(iv)

Explanation

To solve the problem, we need to analyze the oscillation of a rectangular block floating in a liquid. Let's break down the steps:

### Step 1: Understand the forces acting on the block
When the block is floating in equilibrium, the weight of the block (W) is balanced by the buoyant force (FB). The weight of the block is given by:
\[ W = mg \]
where \( m \) is the mass of the block and \( g \) is the acceleration due to gravity.

The buoyant force is given by Archimedes' principle:
\[ F_B = \rho V g \]
where \( \rho \) is the density of the liquid, and \( V \) is the volume of the liquid displaced by the block. For a rectangular block with area of cross-section \( A \) and height \( H \), the volume displaced is:
\[ V = A \cdot h \]
where \( h \) is the height of the submerged part of the block.

### Step 2: Establish equilibrium condition
At equilibrium, the weight of the block equals the buoyant force:
\[ mg = \rho A H g \]
This gives us the relationship between the mass of the block, the area of cross-section, and the density of the liquid.

### Step 3: Analyze small vertical displacements
When the block is displaced vertically by a small distance \( y \), the new submerged volume becomes:
\[ V' = A(H + y) \]
The new buoyant force is:
\[ F_B' = \rho A(H + y)g \]
The change in buoyant force due to the displacement is:
\[ \Delta F_B = F_B' - F_B = \rho A(H + y)g - \rho A Hg = \rho A y g \]

### Step 4: Determine the restoring force
The restoring force acting on the block when it is displaced by \( y \) is given by:
\[ F = -\Delta F_B = -\rho A y g \]

### Step 5: Apply Newton's second law
According to Newton's second law, the acceleration \( a \) of the block can be expressed as:
\[ ma = -\rho A y g \]
This can be rewritten as:
\[ a = -\frac{\rho A g}{m} y \]

### Step 6: Relate to simple harmonic motion
This equation is of the form:
\[ a = -\omega^2 y \]
where \( \omega^2 = \frac{\rho A g}{m} \).

### Step 7: Find the time period \( T \)
The time period \( T \) of oscillation is related to the angular frequency \( \omega \) by:
\[ T = 2\pi \sqrt{\frac{m}{\rho A g}} \]

### Step 8: Analyze the proportional relationships
From the expression for \( T \):
1. \( T \propto \sqrt{m} \) (True)
2. \( T \propto \sqrt{\rho} \) (False)
3. \( T \propto \frac{1}{\sqrt{A}} \) (True)
4. \( T \propto \frac{1}{\sqrt{\rho}} \) (False)

### Conclusion
The correct proportional relationships are:
- \( T \propto \sqrt{m} \)
- \( T \propto \frac{1}{\sqrt{A}} \)

Thus, the correct options are (i) and (iii).