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. Then

Options

A. `T prop sqrt(rho)`
B. `T prop (1)/(sqrt(A))`
C. `T prop (1)/(rho)`
D. `T prop (1)/(sqrt(m))`

Answer

To find the time period \( T \) of oscillation for a rectangular block of mass \( m \) floating in a liquid of density \( \rho \), we can follow these steps:

### Step 1: Understand the equilibrium condition
When the block is floating, it displaces a volume of liquid equal to its weight. The buoyant force acting on the block is equal to the weight of the liquid displaced.

### Step 2: Displace the block
When the block is displaced vertically by a small distance \( x \) from its equilibrium position, the new volume of liquid displaced changes, which affects the buoyant force.

### Step 3: Calculate the change in buoyant force
The buoyant force \( F_b \) acting on the block can be expressed as:
\[ F_b = \rho g V \]
where \( V \) is the volume of the liquid displaced. For a block of cross-sectional area \( A \) and submerged depth \( h \), the volume is \( V = A \cdot h \).

When the block is displaced downward by \( x \), the new submerged depth becomes \( h + x \). Thus, the new buoyant force is:
\[ F_b' = \rho g A (h + x) \]

### Step 4: Determine the restoring force
The restoring force \( F \) acting on the block when it is displaced by \( x \) is given by the difference in buoyant force:
\[ F = F_b' - F_b = \rho g A (h + x) - \rho g A h = \rho g A x \]
This force acts upwards, opposing the downward displacement.

### Step 5: Apply Newton's second law
According to Newton's second law, the net force acting on the block can also be expressed as:
\[ F = m a \]
where \( a \) is the acceleration of the block. Since the restoring force is acting upwards, we can write:
\[ m a = -\rho g A x \]
This can be rearranged to:
\[ a = -\frac{\rho g A}{m} x \]

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

### Step 7: Find the angular frequency \( \omega \)
From the relation \( \omega^2 = \frac{\rho g A}{m} \), we can find \( \omega \):
\[ \omega = \sqrt{\frac{\rho g A}{m}} \]

### Step 8: Calculate the time period \( T \)
The time period \( T \) of oscillation is related to the angular frequency \( \omega \) by:
\[ T = \frac{2\pi}{\omega} \]
Substituting for \( \omega \):
\[ T = 2\pi \sqrt{\frac{m}{\rho g A}} \]

### Conclusion
The time period \( T \) is directly proportional to \( \sqrt{m} \) and inversely proportional to \( \sqrt{\rho} \) and \( \sqrt{A} \).

### Final Result
\[ T \propto \sqrt{m}, \quad T \propto \frac{1}{\sqrt{\rho}}, \quad T \propto \frac{1}{\sqrt{A}} \]