A cylindrical block of the density `rho` is partically immersed in a liwuid of density `3rho`. The plane surface of the block remains parallel to the surface of the liquid. The height of the block is `60 m`. The block perform `SHM` when displaced form its mean position [Use `g = 9.8 m//s^(2)`]

Options

A. the maximum amplitude is `20 cm`.
B. the maximum amplitude is `40 cm`
C. the time period will be `2pi//7` seconds
D. none

Explanation

To solve the problem step by step, we will analyze the forces acting on the cylindrical block and derive the expression for the time period of the simple harmonic motion (SHM) it undergoes when displaced from its equilibrium position.

### Step 1: Understanding the Forces
When the block is partially immersed in the liquid, the buoyant force \( F_b \) acting on the block is equal to the weight of the liquid displaced by the submerged part of the block.

The buoyant force can be expressed as:
\[ F_b = \text{density of liquid} \times \text{volume submerged} \times g \]
Given that the density of the liquid is \( 3\rho \) and the volume submerged when displaced by a distance \( y \) is \( A \cdot y \) (where \( A \) is the cross-sectional area of the block), we have:
\[ F_b = 3\rho \cdot A \cdot y \cdot g \]

### Step 2: Weight of the Block
The weight \( W \) of the block is given by:
\[ W = \text{density of block} \times \text{volume of block} \times g = \rho \cdot A \cdot 60 \cdot g \]

### Step 3: Setting Up the Equation
At equilibrium, the buoyant force equals the weight of the block:
\[ 3\rho \cdot A \cdot 60 \cdot g = \rho \cdot A \cdot 60 \cdot g \]
However, when the block is displaced by \( y \), the new buoyant force becomes:
\[ F_b = 3\rho \cdot A \cdot (60 - y) \cdot g \]
The weight of the block remains the same. The net force \( F \) acting on the block when displaced is:
\[ F = F_b - W = 3\rho \cdot A \cdot (60 - y) \cdot g - \rho \cdot A \cdot 60 \cdot g \]
Simplifying this, we get:
\[ F = 3\rho \cdot A \cdot (60 - y) \cdot g - 3\rho \cdot A \cdot 20 \cdot g \]
\[ F = 3\rho \cdot A \cdot g \cdot (60 - y - 20) = 3\rho \cdot A \cdot g \cdot (40 - y) \]

### Step 4: Restoring Force and SHM
The restoring force \( F \) can be expressed as:
\[ F = -k \cdot y \]
where \( k \) is the effective spring constant. From the previous equation, we can identify:
\[ k = 3\rho \cdot A \cdot g \]
Thus, the equation for SHM becomes:
\[ F = -3\rho \cdot A \cdot g \cdot y \]

### Step 5: Finding the Time Period
The time period \( T \) of SHM is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
where \( m \) is the mass of the block:
\[ m = \rho \cdot A \cdot 60 \]
Substituting for \( k \):
\[ T = 2\pi \sqrt{\frac{\rho \cdot A \cdot 60}{3\rho \cdot A \cdot g}} = 2\pi \sqrt{\frac{60}{3g}} = 2\pi \sqrt{\frac{20}{g}} \]
Substituting \( g = 9.8 \, \text{m/s}^2 \):
\[ T = 2\pi \sqrt{\frac{20}{9.8}} = 2\pi \sqrt{\frac{100}{49}} = 2\pi \cdot \frac{10}{7} = \frac{20\pi}{7} \]

### Final Answer
Thus, the time period \( T \) of the SHM is:
\[ T = \frac{20\pi}{7} \, \text{s} \]