A block of mass `5` kg and volume `0.05"m"^(3)` floats in a liquid of density `140 "kg m"^(-3)` .
Fraction of block outside liquid is

Options

A. `(2)/(7)`
B. `(7)/(2)`
C. `(3)/(7)`
D. `(7)/(3)`

Explanation

To find the fraction of the block that is outside the liquid, we can follow these steps:

### Step 1: Write down the given data
- Mass of the block, \( m = 5 \, \text{kg} \)
- Volume of the block, \( V = 0.05 \, \text{m}^3 \)
- Density of the liquid, \( \rho = 140 \, \text{kg/m}^3 \)

### Step 2: Calculate the volume of the block submerged in the liquid
Using the principle of buoyancy, the weight of the block is equal to the weight of the liquid displaced by the submerged part of the block.

The weight of the block is given by:
\[ W = m \cdot g \]
Where \( g \) is the acceleration due to gravity. Since \( g \) will cancel out later, we can ignore it for now.

The weight of the displaced liquid can be expressed as:
\[ W_{\text{displaced}} = V_i \cdot \rho \cdot g \]
Where \( V_i \) is the volume of the block submerged in the liquid.

Setting these two weights equal gives us:
\[ m = V_i \cdot \rho \]
Thus, we can find the volume submerged:
\[ V_i = \frac{m}{\rho} \]
Substituting the values:
\[ V_i = \frac{5 \, \text{kg}}{140 \, \text{kg/m}^3} = \frac{5}{140} = \frac{1}{28} \, \text{m}^3 \]

### Step 3: Calculate the volume of the block outside the liquid
The volume of the block outside the liquid, \( V_0 \), can be calculated as:
\[ V_0 = V - V_i \]
Substituting the values:
\[ V_0 = 0.05 \, \text{m}^3 - \frac{1}{28} \, \text{m}^3 \]

To perform this subtraction, we need a common denominator. The volume \( 0.05 \, \text{m}^3 \) can be expressed as:
\[ 0.05 = \frac{5}{100} = \frac{1}{20} \]
Finding a common denominator (which is 140):
\[ \frac{1}{20} = \frac{7}{140} \quad \text{and} \quad \frac{1}{28} = \frac{5}{140} \]
Now, substituting these values:
\[ V_0 = \frac{7}{140} - \frac{5}{140} = \frac{2}{140} = \frac{1}{70} \, \text{m}^3 \]

### Step 4: Calculate the fraction of the block outside the liquid
The fraction \( f \) of the block outside the liquid is given by:
\[ f = \frac{V_0}{V} \]
Substituting the values:
\[ f = \frac{\frac{1}{70}}{0.05} \]
Expressing \( 0.05 \) as \( \frac{5}{100} = \frac{1}{20} \):
\[ f = \frac{1}{70} \times \frac{20}{1} = \frac{20}{70} = \frac{2}{7} \]

### Final Answer
The fraction of the block outside the liquid is:
\[ \frac{2}{7} \]

---