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

Options

A. `0.024 "m"^(3)`
B. `0.014"m"^(3)`
C. `0.015"m"^(3)`
D. `0.041"m"^(3)`

Explanation

To solve the problem of finding the volume of the block above the surface of the liquid, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Mass of the block, \( m = 5 \, \text{kg} \)
- Volume of the block, \( V = 0.05 \, \text{m}^3 \)
- Density of the liquid, \( \rho = 140 \, \text{kg/m}^3 \)

2. **Calculate the Weight of the Block**:
The weight of the block can be calculated using the formula:
\[ W = m \cdot g \]
where \( g \) (acceleration due to gravity) is approximately \( 9.81 \, \text{m/s}^2 \). However, for this problem, we can ignore \( g \) since it will cancel out later.

3. **Apply the Principle of Buoyancy**:
According to Archimedes' principle, the weight of the liquid displaced by the block (upthrust) is equal to the weight of the block when it is floating. The upthrust can be expressed as:
\[ U = V_i \cdot \rho \cdot g \]
where \( V_i \) is the volume of the block submerged in the liquid.

4. **Set Up the Equation**:
Since the block is floating in equilibrium:
\[ W = U \]
Therefore,
\[ m \cdot g = V_i \cdot \rho \cdot g \]
We can cancel \( g \) from both sides:
\[ m = V_i \cdot \rho \]

5. **Solve for the Volume Submerged**:
Rearranging the equation gives:
\[ V_i = \frac{m}{\rho} \]
Substituting the known values:
\[ V_i = \frac{5 \, \text{kg}}{140 \, \text{kg/m}^3} = \frac{5}{140} = \frac{1}{28} \, \text{m}^3 \]

6. **Calculate the Volume Above the Surface**:
The volume of the block above the surface of the liquid, \( V_o \), can be found by subtracting the volume submerged from the total volume of the block:
\[ V_o = V - V_i \]
Substituting the values:
\[ V_o = 0.05 \, \text{m}^3 - \frac{1}{28} \, \text{m}^3 \]

7. **Convert \( 0.05 \, \text{m}^3 \) to a Fraction**:
Convert \( 0.05 \) to a fraction:
\[ 0.05 = \frac{5}{100} = \frac{1}{20} \]

8. **Find a Common Denominator and Calculate**:
The common denominator of 20 and 28 is 140. Convert both fractions:
\[ V_o = \frac{7}{140} - \frac{5}{140} = \frac{2}{140} = \frac{1}{70} \, \text{m}^3 \]

9. **Final Calculation**:
Converting \( \frac{1}{70} \) to decimal gives:
\[ V_o \approx 0.0142857 \, \text{m}^3 \]

### Final Answer:
The volume of the block above the surface of the liquid is approximately:
\[ \boxed{0.014 \, \text{m}^3} \]