A cylindrical of wood (density `= 600 kg m^(-3)`) of base area `30 cm^(2)` and height `54 cm`, floats in a liquid of density `900 kg^(-3)` The block is deapressed slightly and then released. The time period of the resulting oscillations of the block would be equal to that of a simple pendulum of length (nearly) :

Options

A. `26 cm`
B. `52 cm`
C. `36cm`
D. `65 cm`

Explanation

To solve the problem step by step, we need to determine the time period of oscillation of a cylindrical wooden block floating in a liquid. The key is to relate the oscillation of the block to that of a simple pendulum.

### Step 1: Understand the System
The wooden block is floating in a liquid, and its weight is balanced by the buoyant force when it is in equilibrium. When the block is slightly depressed and then released, it will oscillate.

### Step 2: Identify the Parameters
- Density of wood, \( \rho_s = 600 \, \text{kg/m}^3 \)
- Density of liquid, \( \rho_l = 900 \, \text{kg/m}^3 \)
- Base area of the cylinder, \( A = 30 \, \text{cm}^2 = 30 \times 10^{-4} \, \text{m}^2 = 3 \times 10^{-3} \, \text{m}^2 \)
- Height of the cylinder, \( h = 54 \, \text{cm} = 0.54 \, \text{m} \)

### Step 3: Calculate the Volume of the Cylinder
The volume \( V \) of the cylinder is given by:
\[ V = A \cdot h = (3 \times 10^{-3} \, \text{m}^2) \cdot (0.54 \, \text{m}) = 1.62 \times 10^{-3} \, \text{m}^3 \]

### Step 4: Calculate the Weight of the Wooden Block
The weight \( W \) of the wooden block is:
\[ W = \rho_s \cdot V \cdot g = 600 \, \text{kg/m}^3 \cdot 1.62 \times 10^{-3} \, \text{m}^3 \cdot 9.81 \, \text{m/s}^2 \]
Calculating this gives:
\[ W = 600 \cdot 1.62 \cdot 9.81 \approx 9526.92 \, \text{N} \]

### Step 5: Calculate the Buoyant Force
The buoyant force \( F_b \) when the block is submerged to a depth \( x \) is given by:
\[ F_b = \rho_l \cdot V_{\text{submerged}} \cdot g = \rho_l \cdot (A \cdot x) \cdot g \]
When the block is displaced by \( x \), the increase in buoyant force is:
\[ \Delta F_b = \rho_l \cdot A \cdot g \cdot x \]

### Step 6: Establish the Restoring Force
The restoring force \( F \) is equal to the increase in buoyant force:
\[ F = \Delta F_b = \rho_l \cdot A \cdot g \cdot x \]

### Step 7: Relate to Simple Harmonic Motion
The time period \( T \) of the oscillation can be related to the mass \( m \) of the block and the effective spring constant \( k \):
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
Here, \( k = \rho_l \cdot A \cdot g \) and \( m = \rho_s \cdot V \).

### Step 8: Substitute Values
Substituting the values, we find:
\[ T = 2\pi \sqrt{\frac{\rho_s \cdot V}{\rho_l \cdot A \cdot g}} \]

### Step 9: Calculate the Length of the Equivalent Pendulum
We want to find the length \( L \) of the equivalent simple pendulum:
\[ L = \frac{\rho_s}{\rho_l} \cdot h \]
Substituting the values:
\[ L = \frac{600}{900} \cdot 54 \, \text{cm} = \frac{2}{3} \cdot 54 \, \text{cm} = 36 \, \text{cm} \]

### Final Answer
The time period of the resulting oscillations of the block would be equal to that of a simple pendulum of length approximately **36 cm**.