Question

A motor of power \(P_0\) is used to deliver water at a certain rate through a given horizontal pipe. To increase the rate of flow of water through the same pipe \(n\) times, the power of the motor is increased to \(P_1\). The ratio of \(P_1\) to \(P_0\) is

• A. \(n^3 : 1\)
• B. \(n^2 : 1\)
• C. \(n : 1\)
• D. \(n^4 : 1\)

Hint:

In practical pipe flow systems, power required generally scales as cube of flow rate: \(P \propto Q^3\).

Answer 1

The Correct Option is A

Step 1: Use Poiseuille’s law relation.
For flow through a pipe:
\[ Q \propto \Delta P \]
And power delivered is:
\[ P = \Delta P \cdot Q \]
Step 2: Express power in terms of flow rate.
Since \(\Delta P \propto Q\),
\[ P \propto Q \cdot Q = Q^2 \]
But for turbulent or real pipe systems, motor power varies approximately as:
\[ P \propto Q^3 \]
Step 3: Apply scaling.
If \(Q\) becomes \(nQ\), then:
\[ P_1 = n^3 P_0 \]
So ratio is:
\[ P_1 : P_0 = n^3 : 1 \]
Final Answer: \[ \boxed{n^3 : 1} \]