A pump motor is used to deliver water at a certain rate from a given pipe. To obtain, twice as much water from the same pipe, in the same time, the power of motor has to be increased to n times. Find n.

Answer

To solve the problem, we need to determine how much the power of the pump motor must be increased to deliver twice the amount of water in the same time. Let's break this down step by step.

### Step 1: Understand the relationship between power, energy, and time
Power (P) is defined as the rate at which work is done or energy is transferred. Mathematically, it can be expressed as:
\[ P = \frac{E}{t} \]
where \( E \) is the energy and \( t \) is the time.

### Step 2: Calculate the initial power
Assume the initial velocity of water being pumped is \( v \) and the mass of water delivered in time \( t \) is \( m \). The kinetic energy (E) of the water being pumped can be expressed as:
\[ E = \frac{1}{2} mv^2 \]
Thus, the initial power \( P \) can be calculated as:
\[ P = \frac{E}{t} = \frac{\frac{1}{2} mv^2}{t} = \frac{mv^2}{2t} \]

### Step 3: Calculate the new power for twice the water
To deliver twice as much water in the same time, the mass of water delivered becomes \( 2m \). The new velocity required to achieve this is \( 2v \). The new kinetic energy (E') can be expressed as:
\[ E' = \frac{1}{2} (2m)(2v)^2 = \frac{1}{2} (2m)(4v^2) = 4mv^2 \]
Now, the new power \( P' \) can be calculated as:
\[ P' = \frac{E'}{t} = \frac{4mv^2}{t} \]

### Step 4: Relate the new power to the initial power
From the expressions for power, we have:
- Initial power: \( P = \frac{mv^2}{2t} \)
- New power: \( P' = \frac{4mv^2}{t} \)

Now, we can express \( P' \) in terms of \( P \):
\[ P' = 4 \left( \frac{mv^2}{2t} \right) = 4P \]

### Step 5: Determine the factor of increase in power
We know from the problem statement that:
\[ P' = nP \]
From our calculations, we found that:
\[ P' = 4P \]
Thus, we can equate:
\[ nP = 4P \]
This implies:
\[ n = 4 \]

### Conclusion
The power of the motor must be increased by a factor of \( n = 4 \) to deliver twice as much water from the same pipe in the same time.