An engine pumps water through a hose pipe. Water passes through the pipe and leaves it with a velocity of `2 ms^(1)`. The mass per unit length of water in the pipe is `100 kgm^(-1)`. What is the power of the engine?
(a) 400 W (b) 200W (c) 100W (d) 800W

Options

A. 400 W
B. 200W
C. 100W
D. 800W

Answer

To solve the problem of finding the power of the engine that pumps water through a hose pipe, we can follow these steps:

### Step 1: Understand the relationship between power, force, and velocity.
Power (P) can be expressed as the product of force (F) and velocity (v):
\[ P = F \cdot v \]
This is our equation (1).

### Step 2: Determine the force exerted by the water.
The force can be defined as the change in momentum per unit time. Since the water is being pumped with a constant velocity, we can express the force in terms of mass flow rate. The mass flow rate (ṁ) can be defined as:
\[ \dot{m} = \frac{m}{t} \]
Where \( m \) is the mass and \( t \) is the time.

### Step 3: Relate mass to mass per unit length.
Given that the mass per unit length of water in the pipe is \( 100 \, \text{kg/m} \), we can express the mass of water in terms of the length of the pipe (L):
\[ m = 100 \cdot L \]

### Step 4: Calculate the mass flow rate.
The volume flow rate (Q) can be expressed as:
\[ Q = A \cdot v \]
Where \( A \) is the cross-sectional area of the pipe and \( v \) is the velocity of the water. The mass flow rate can then be expressed as:
\[ \dot{m} = \rho \cdot Q = \rho \cdot A \cdot v \]
Since we are given the mass per unit length, we can relate it to the density (ρ) and area (A).

### Step 5: Substitute into the power equation.
Now, substituting the mass flow rate into the power equation:
\[ P = \dot{m} \cdot v = (100 \cdot v) \cdot v = 100 \cdot v^2 \]
Where \( v = 2 \, \text{m/s} \).

### Step 6: Calculate the power.
Substituting the value of \( v \):
\[ P = 100 \cdot (2)^2 = 100 \cdot 4 = 400 \, \text{W} \]

### Final Answer:
The power of the engine is \( 400 \, \text{W} \), which corresponds to option (a).

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