Two unequal blocks placed over each other of densities `sigma_(1)` and `sigma_(2)` are immersed in a fluid of density of `sigma`. The block of density `sigma_(1)` is fully submerged so that ratio of their masses is `1/2` and `sigma//sigma_(1)=2` and `sigma//sigma_(2)=0.5`. Find the degree of submergence of the upper block of density `sigma_(2)`.

Options

A. `50%` submerged
B. `25%` submerged
C. `75%` submerged
D. fully submerged

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have two blocks with densities \( \sigma_1 \) and \( \sigma_2 \) placed over each other. The block with density \( \sigma_1 \) is fully submerged in a fluid of density \( \sigma \). Given the ratios of their masses and the densities, we need to find the degree of submergence of the upper block with density \( \sigma_2 \).

### Step 2: Given Information
1. The ratio of the masses of the two blocks is \( \frac{m_1}{m_2} = \frac{1}{2} \).
2. The density ratios are given as:
- \( \frac{\sigma}{\sigma_1} = 2 \) which implies \( \sigma_1 = \frac{\sigma}{2} \).
- \( \frac{\sigma}{\sigma_2} = 0.5 \) which implies \( \sigma_2 = 2\sigma \).

### Step 3: Express Masses in Terms of Volume
Let the volumes of the blocks be \( V_1 \) for the block with density \( \sigma_1 \) and \( V_2 \) for the block with density \( \sigma_2 \).

- The mass of block 1:
\[ m_1 = \sigma_1 V_1 = \left(\frac{\sigma}{2}\right) V_1 \]
- The mass of block 2:
\[ m_2 = \sigma_2 V_2 = (2\sigma) V_2 \]

### Step 4: Use the Mass Ratio
From the mass ratio \( \frac{m_1}{m_2} = \frac{1}{2} \):
\[ \frac{\frac{\sigma}{2} V_1}{2\sigma V_2} = \frac{1}{2} \]
Cross-multiplying gives:
\[ \sigma V_1 = 2 \cdot 2\sigma V_2 \implies V_1 = 4 V_2 \]

### Step 5: Buoyant Force and Weight Balance
The block with density \( \sigma_1 \) is fully submerged, so the buoyant force acting on it is:
\[ F_b = \sigma V_1 g \]
The total weight of both blocks is:
\[ W = m_1 g + m_2 g = \left(\frac{\sigma}{2} V_1 + 2\sigma V_2\right) g \]

### Step 6: Set Up the Equation
Setting the total weight equal to the buoyant force:
\[ \sigma V_1 = \left(\frac{\sigma}{2} V_1 + 2\sigma V_2\right) \]
Dividing through by \( \sigma \) (assuming \( \sigma \neq 0 \)):
\[ V_1 = \frac{1}{2} V_1 + 2 V_2 \]
Rearranging gives:
\[ \frac{1}{2} V_1 = 2 V_2 \implies V_1 = 4 V_2 \]
This confirms our earlier finding.

### Step 7: Degree of Submergence of Block 2
Let \( x \) be the fraction of the volume of block 2 that is submerged. The volume of block 2 is \( V_2 \), so the volume submerged is \( x V_2 \).

The buoyant force on block 2 is:
\[ F_b = \sigma (x V_2) g \]
Setting the total weight equal to the buoyant force:
\[ \left(2\sigma V_2\right) g = \sigma (x V_2) g \]
Cancelling \( g \) and \( \sigma \) (assuming \( \sigma \neq 0 \)):
\[ 2 V_2 = x V_2 \]
Dividing both sides by \( V_2 \) (assuming \( V_2 \neq 0 \)):
\[ 2 = x \]
Thus, \( x = 2 \), which means the upper block is fully submerged.

### Conclusion
The degree of submergence of the upper block of density \( \sigma_2 \) is 100% or fully submerged.

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