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 situation involving the cylindrical block and the liquid.

### Step 1: Understand the System
We have a cylindrical block of density \( \rho \) partially immersed in a liquid of density \( 3\rho \). The height of the block is \( 60 \, \text{cm} \) (or \( 0.6 \, \text{m} \)). The block performs simple harmonic motion (SHM) when displaced from its mean position.

### Step 2: Determine the Equilibrium Condition
At equilibrium, the buoyant force acting on the block equals the weight of the block. The buoyant force \( F_b \) can be expressed as:
\[ F_b = \text{density of liquid} \times \text{volume of liquid displaced} \times g \]
The volume of liquid displaced when the block is partially submerged to a height \( y \) is:
\[ \text{Volume displaced} = A \cdot y \]
where \( A \) is the cross-sectional area of the block.

Thus, the buoyant force becomes:
\[ F_b = 3\rho \cdot (A \cdot y) \cdot g \]

The weight of the block \( W \) is:
\[ W = \text{density of block} \times \text{volume of block} \times g = \rho \cdot (A \cdot 0.6) \cdot g \]

Setting the buoyant force equal to the weight of the block gives us:
\[ 3\rho \cdot (A \cdot y) \cdot g = \rho \cdot (A \cdot 0.6) \cdot g \]

### Step 3: Simplify the Equation
We can cancel \( A \) and \( g \) from both sides:
\[ 3y = 0.6 \]
Solving for \( y \):
\[ y = \frac{0.6}{3} = 0.2 \, \text{m} = 20 \, \text{cm} \]

### Step 4: Determine the Amplitude of SHM
The amplitude of the SHM is the maximum displacement from the equilibrium position. Since the block is initially at equilibrium with \( y = 20 \, \text{cm} \), this is the amplitude of the motion.

### Step 5: Calculate the Time Period of SHM
The effective spring constant \( k \) for the system can be derived from the buoyant force. The restoring force when the block is displaced by a small distance \( x \) from its equilibrium position is:
\[ F = -\Delta V \cdot \text{density of liquid} \cdot g = -3\rho A (y + x) g + 3\rho A y g \]
This simplifies to:
\[ F = -3\rho A g \cdot x \]
Thus, \( k = 3\rho A g \).

The mass \( m \) of the block is:
\[ m = \rho \cdot (A \cdot 0.6) \]

Using the formula for the time period \( T \) of SHM:
\[ T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{\rho \cdot (A \cdot 0.6)}{3\rho A g}} = 2\pi \sqrt{\frac{0.6}{3g}} \]

### Step 6: Substitute Values
Substituting \( g = 9.8 \, \text{m/s}^2 \):
\[ T = 2\pi \sqrt{\frac{0.6}{3 \cdot 9.8}} = 2\pi \sqrt{\frac{0.6}{29.4}} = 2\pi \sqrt{0.0204} \]
Calculating \( T \):
\[ T \approx 2\pi \cdot 0.142 = 0.891 \, \text{s} \approx 2\pi / 7 \, \text{s} \]

### Final Results
- The maximum amplitude of the SHM is \( 20 \, \text{cm} \).
- The time period of the SHM is approximately \( \frac{2\pi}{7} \, \text{s} \).