A liquid of specific gravity `0.8` is flowing in a pipe line with a speed of `2 m//s`. The `K.E` per cubic meter of it is

Options

A. 160 J
B. 1600 J
C. 160.5 J
D. 1.6 J

Explanation

To find the kinetic energy (K.E) per cubic meter of a liquid flowing in a pipeline, we can follow these steps:

### Step 1: Understand the formula for kinetic energy
The kinetic energy (K.E) of a fluid can be expressed as:
\[ K.E = \frac{1}{2} m v^2 \]
where:
- \( m \) is the mass of the fluid,
- \( v \) is the velocity of the fluid.

### Step 2: Relate mass to volume
Since we are interested in kinetic energy per cubic meter, we can express mass in terms of density:
\[ m = \rho \cdot V \]
where:
- \( \rho \) is the density of the liquid,
- \( V \) is the volume (1 cubic meter in this case).

Thus, the kinetic energy per cubic meter can be written as:
\[ K.E = \frac{1}{2} \rho V v^2 \]
Dividing by the volume \( V \) (which is 1 cubic meter), we have:
\[ K.E \text{ per cubic meter} = \frac{1}{2} \rho v^2 \]

### Step 3: Find the density of the liquid
The specific gravity of the liquid is given as \( 0.8 \). Specific gravity is the ratio of the density of a substance to the density of water. The density of water is approximately \( 1000 \, \text{kg/m}^3 \).
Thus, the density of the liquid can be calculated as:
\[ \rho = \text{Specific Gravity} \times \text{Density of Water} \]
\[ \rho = 0.8 \times 1000 \, \text{kg/m}^3 = 800 \, \text{kg/m}^3 \]

### Step 4: Substitute the values into the kinetic energy formula
Now we can substitute the values of \( \rho \) and \( v \) into the kinetic energy formula:
\[ K.E \text{ per cubic meter} = \frac{1}{2} \rho v^2 \]
\[ K.E = \frac{1}{2} \times 800 \, \text{kg/m}^3 \times (2 \, \text{m/s})^2 \]

### Step 5: Calculate the kinetic energy
Calculating further:
\[ K.E = \frac{1}{2} \times 800 \times 4 \]
\[ K.E = 400 \times 4 \]
\[ K.E = 1600 \, \text{Joules/m}^3 \]

### Conclusion
The kinetic energy per cubic meter of the liquid flowing in the pipeline is \( 1600 \, \text{Joules} \).

---