A pump is used to pump a liquid of density `rho` continuously through a pipe of cross, sectional area A. If liquid is flowing with speed v, then average power of pump is

Options

A. `1/3rhoAV^(2)`
B. `1/2rhoAV^(2)`
C. `2rhoAV^(2)`
D. `1/2rhoAV^(3)`

Answer

To find the average power of the pump that is used to pump a liquid of density \( \rho \) through a pipe of cross-sectional area \( A \) at a speed \( v \), we can follow these steps:

### Step 1: Understand the definition of power
Power is defined as the rate at which work is done or energy is transferred. In this case, we will calculate the average power of the pump based on the energy required to move the liquid.

### Step 2: Calculate the kinetic energy of the liquid
The kinetic energy (KE) of the liquid flowing through the pipe can be expressed as:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the liquid and \( v \) is its speed.

### Step 3: Relate mass to density and volume
The mass \( m \) of the liquid can be expressed in terms of its density \( \rho \) and volume \( V \):
\[ m = \rho V \]
where \( V \) is the volume of the liquid flowing through the pipe.

### Step 4: Determine the volume flow rate
The volume flow rate \( Q \) can be defined as:
\[ Q = A v \]
where \( A \) is the cross-sectional area of the pipe. This means that the volume of liquid flowing in a time \( t \) is:
\[ V = Q \cdot t = A v t \]

### Step 5: Substitute the volume into the kinetic energy formula
Now, substituting the expression for volume \( V \) into the kinetic energy formula:
\[ KE = \frac{1}{2} (\rho A v t) v^2 = \frac{1}{2} \rho A v^3 t \]

### Step 6: Calculate the average power
The average power \( P \) is the energy per unit time. Therefore, we can express the average power as:
\[ P = \frac{KE}{t} = \frac{\frac{1}{2} \rho A v^3 t}{t} = \frac{1}{2} \rho A v^3 \]

### Final Result
Thus, the average power of the pump is:
\[ P = \frac{1}{2} \rho A v^3 \]