A block of ice of total area A and thicknes `0.5m` is floating in water. In order to just support a man of nass `100kg`, the area A should be (the specific gravity of ice is 0.9)

Options

A. `2.2m^(2)`
B. `1.0m^(2)`
C. `0.02m^(2)`
D. None of these

Explanation

To solve the problem of determining the area \( A \) of the block of ice required to support a man of mass \( 100 \, \text{kg} \), we can follow these steps:

### Step 1: Understand the Forces Acting on the Ice
When the ice block is floating, the weight of the man and the weight of the ice block must equal the buoyant force (upthrust) exerted by the water.

### Step 2: Write the Equation for Equilibrium
The equilibrium condition can be expressed as:
\[ \text{Weight of the man} + \text{Weight of the ice} = \text{Buoyant force} \]
Mathematically, this can be written as:
\[ m_1 g + m_2 g = V \rho_{\text{water}} g \]
where:
- \( m_1 = 100 \, \text{kg} \) (mass of the man)
- \( m_2 \) is the mass of the ice
- \( V \) is the volume of the displaced water (which is equal to the volume of the ice block submerged)
- \( \rho_{\text{water}} = 1000 \, \text{kg/m}^3 \) (density of water)
- \( g \) is the acceleration due to gravity (which will cancel out)

### Step 3: Calculate the Mass of the Ice Block
The mass of the ice block can be calculated using its volume and density. The specific gravity of ice is given as \( 0.9 \), which means:
\[ \rho_{\text{ice}} = 0.9 \times \rho_{\text{water}} = 0.9 \times 1000 \, \text{kg/m}^3 = 900 \, \text{kg/m}^3 \]
The volume \( V \) of the ice block can be expressed in terms of its area \( A \) and thickness \( t \):
\[ V = A \times t \]
Given that the thickness \( t = 0.5 \, \text{m} \), we can express the mass of the ice block as:
\[ m_2 = V \rho_{\text{ice}} = (A \times 0.5) \times 900 \]

### Step 4: Substitute and Rearrange the Equation
Substituting \( m_2 \) into the equilibrium equation:
\[ 100 \, \text{kg} \cdot g + (A \times 0.5 \times 900) \cdot g = (A \times 0.5) \times 1000 \cdot g \]
Cancelling \( g \) from all terms:
\[ 100 + 450A = 500A \]

### Step 5: Solve for Area \( A \)
Rearranging the equation gives:
\[ 100 = 500A - 450A \]
\[ 100 = 50A \]
\[ A = \frac{100}{50} = 2 \, \text{m}^2 \]

### Conclusion
The area \( A \) required to just support the man is \( 2 \, \text{m}^2 \).