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. `Tprop(1)/(sqrt(m))`
B. `Tpropsqrt(rho)`
C. `Tprop(1)/(sqrt(A))`
D. `Tprop(1)/(rho)`

Explanation

To solve the problem regarding the time period of oscillation of a rectangular block floating in a liquid, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a rectangular block of mass \( m \) and area of cross-section \( A \) floating in a liquid of density \( \rho \).
- When the block is displaced vertically by a small distance \( x \) from its equilibrium position, it will experience a restoring force due to buoyancy.

2. **Buoyancy Force**:
- The buoyant force acting on the block when it is submerged is given by Archimedes' principle:
\[ F_B = \rho g V \]
where \( V \) is the volume of the liquid displaced. For a rectangular block, if it is displaced by \( x \), the volume displaced is \( A \cdot x \). Thus, the buoyant force becomes:
\[ F_B = \rho g (A \cdot x) \]

3. **Restoring Force**:
- The restoring force \( F_R \) acting on the block when it is displaced is equal to the buoyant force:
\[ F_R = -\rho g A x \]
- The negative sign indicates that the force acts in the opposite direction of the displacement.

4. **Equation of Motion**:
- According to Hooke's law for simple harmonic motion (SHM), the restoring force can also be expressed as:
\[ F_R = -m \omega^2 x \]
- Setting the two expressions for restoring force equal gives:
\[ -m \omega^2 x = -\rho g A x \]
- We can cancel \( x \) (assuming \( x \neq 0 \)):
\[ m \omega^2 = \rho g A \]

5. **Finding Angular Frequency**:
- Rearranging the equation gives:
\[ \omega^2 = \frac{\rho g A}{m} \]

6. **Time Period of Oscillation**:
- The time period \( T \) of oscillation is related to the angular frequency \( \omega \) by the formula:
\[ T = \frac{2\pi}{\omega} \]
- Substituting for \( \omega \):
\[ T = 2\pi \sqrt{\frac{m}{\rho g A}} \]

### Final Result:
The time period \( T \) of the oscillation of the block is given by:
\[ T = 2\pi \sqrt{\frac{m}{\rho g A}} \]