A pump motor is used to deliver water at a certain rate from a given pipe. To obtain 'n' times water from the same pipe in the same time by what amount (a) the force and (b) power of the motor should be increased ?

Options

A. (a) `n^2` times
B. (b) `n^3` times
C. (c) `n` times
D. (d) `n^(3//2)` times

Explanation

To solve the problem, we need to determine how much the force and power of the pump motor should be increased to deliver 'n' times the amount of water in the same time. Let's break it down step by step.

### Step 1: Understand the mass flow rate
The mass flow rate of water being delivered by the pump can be expressed as:
\[ \frac{dm}{dt} = \rho \cdot A \cdot V \]
where:
- \(\rho\) = density of water
- \(A\) = cross-sectional area of the pipe
- \(V\) = velocity of water

### Step 2: Determine the new mass flow rate
To obtain 'n' times the water in the same time, the new mass flow rate (\(\frac{dm}{dt}'\)) will be:
\[ \frac{dm}{dt}' = n \cdot \frac{dm}{dt} \]

### Step 3: Relate the new parameters
Since the density of water (\(\rho\)) and the area of the pipe (\(A\)) remain constant, we can write:
\[ \rho \cdot A \cdot V' = n \cdot (\rho \cdot A \cdot V) \]
This simplifies to:
\[ V' = n \cdot V \]
indicating that the new velocity of water (\(V'\)) must be 'n' times the original velocity (\(V\)).

### Step 4: Calculate the force
The force exerted by the pump can be defined as:
\[ F = V \cdot \frac{dm}{dt} \]
To find the new force (\(F'\)):
\[ F' = V' \cdot \frac{dm}{dt}' = (n \cdot V) \cdot (n \cdot \frac{dm}{dt}) = n^2 \cdot F \]
Thus, the force must be increased by a factor of \(n^2\).

### Step 5: Calculate the power
Power can be defined as:
\[ P = F \cdot V \]
For the new power (\(P'\)):
\[ P' = F' \cdot V' = (n^2 \cdot F) \cdot (n \cdot V) = n^3 \cdot (F \cdot V) = n^3 \cdot P \]
Thus, the power must be increased by a factor of \(n^3\).

### Final Answers
(a) The force should be increased by \(n^2\) times.
(b) The power should be increased by \(n^3\) times.

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