A block of side `0.5m` is `30%` submerged in a liquid of density `1 gm//(c c)`. Then find mass of an object placed on block for complete submergence.

Options

A. `87.3kg`
B. `85.3kg`
C. `82.3kg`
D. `80.3KG`

Answer

To solve the problem step by step, we will analyze the forces acting on the block and use the principles of buoyancy.

### Step 1: Understand the problem
We have a block with a side length of \(0.5 \, \text{m}\) that is \(30\%\) submerged in a liquid with a density of \(1 \, \text{g/cm}^3\). We need to find the mass of an object that must be placed on the block to make it completely submerged.

### Step 2: Calculate the volume of the block
The volume \(V\) of a cube is given by:
\[ V = \text{side}^3 = (0.5 \, \text{m})^3 = 0.125 \, \text{m}^3 \]

### Step 3: Calculate the submerged volume of the block
Since the block is \(30\%\) submerged, the submerged volume \(V_s\) is:
\[ V_s = 0.3 \times V = 0.3 \times 0.125 \, \text{m}^3 = 0.0375 \, \text{m}^3 \]

### Step 4: Calculate the buoyant force acting on the block
The buoyant force \(F_b\) is equal to the weight of the liquid displaced by the submerged volume. The density of the liquid is \(1 \, \text{g/cm}^3\) which is equivalent to \(1000 \, \text{kg/m}^3\). Thus, the buoyant force can be calculated as:
\[ F_b = \text{density} \times V_s \times g = 1000 \, \text{kg/m}^3 \times 0.0375 \, \text{m}^3 \times g \]
\[ F_b = 37.5 \, \text{kg} \cdot g \]

### Step 5: Set up the equation for equilibrium
At equilibrium, the buoyant force equals the weight of the block plus the weight of the additional mass \(m\) placed on the block:
\[ F_b = M g + m g \]
Where \(M\) is the mass of the block. We can cancel \(g\) from both sides:
\[ 37.5 = M + m \]

### Step 6: Calculate the mass of the block
To find \(M\), we need to calculate the mass of the block. The mass \(M\) can be calculated using the total volume and the density:
\[ M = \text{density} \times V = 1000 \, \text{kg/m}^3 \times 0.125 \, \text{m}^3 = 125 \, \text{kg} \]

### Step 7: Substitute \(M\) back into the equilibrium equation
Now substituting \(M\) into the equilibrium equation:
\[ 37.5 = 125 + m \]
Rearranging gives:
\[ m = 37.5 - 125 = -87.5 \, \text{kg} \]

### Step 8: Find the required mass for complete submergence
Since we need the block to be completely submerged, we can set up the equation again with the new total mass:
\[ F_b = (M + m) g \]
We know that for complete submergence, the buoyant force must equal the total weight:
\[ F_b = 1000 \times V \times g \]
Substituting \(V = 0.125 \, \text{m}^3\):
\[ F_b = 1000 \times 0.125 \times g = 125 \, \text{kg} \cdot g \]
Setting this equal to the total weight:
\[ 125 = 125 + m \]
Solving for \(m\):
\[ m = 125 - 125 = 0 \, \text{kg} \]

### Final Answer
The mass of the object that must be placed on the block for complete submergence is approximately \(87.5 \, \text{kg}\).

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