An engine pumps liquid of density ρ continuously through a pipe of cross-section area A. If the speed with which liquid passes through the pipe is v, then the rate at which kinetic energy is being imparted to the liquid by the pump is

Options

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

Explanation

To find the rate at which kinetic energy is being imparted to the liquid by the pump, we can follow these steps:

### Step 1: Understand the Kinetic Energy Formula
The kinetic energy (KE) of an object with mass \( m \) moving at a speed \( v \) is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]

### Step 2: Determine the Mass Flow Rate
The mass flow rate (\( \dot{m} \)) of the liquid can be expressed in terms of its density (\( \rho \)), cross-sectional area (\( A \)), and velocity (\( v \)):
\[ \dot{m} = \rho \cdot Q \]
where \( Q \) is the volumetric flow rate. The volumetric flow rate can be expressed as:
\[ Q = A \cdot v \]
Thus, the mass flow rate becomes:
\[ \dot{m} = \rho \cdot (A \cdot v) = \rho A v \]

### Step 3: Calculate the Rate of Kinetic Energy
The rate at which kinetic energy is imparted to the liquid (power, \( P \)) is the kinetic energy per unit time:
\[ P = \frac{1}{2} \dot{m} v^2 \]
Substituting the expression for \( \dot{m} \):
\[ P = \frac{1}{2} (\rho A v) v^2 \]

### Step 4: Simplify the Expression
Now, simplify the expression:
\[ P = \frac{1}{2} \rho A v^3 \]

### Final Answer
Thus, the rate at which kinetic energy is being imparted to the liquid by the pump is:
\[ \boxed{\frac{1}{2} \rho A v^3} \]

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