Question

A pump is used to deliver water at a certain rate from a given pipe. To obtain twice the volume of water from the same pipe in the same time, by what factor must the power of the motor pump be increased? Assume ideal conditions.

• A. \(4\)
• B. \(8\)
• C. \(16\)
• D. \(32\)

Hint:

For fluid flow through the same pipe, if flow speed becomes \(n\) times, power becomes \(n^3\) times.

Answer 1

The Correct Option is B

Concept: For water flowing through the same pipe, the volume flow rate is proportional to velocity: \[ Q=Av \] where \(A\) is the area of cross-section and \(v\) is the velocity of water. Power required to maintain the flow is proportional to: \[ P\propto v^3 \]

Step 1: Since the pipe is the same, area \(A\) remains constant. If volume of water delivered in the same time is doubled, then flow rate doubles: \[ Q' = 2Q \] Since: \[ Q=Av \] and \(A\) is constant, \[ v'=2v \]

Step 2: Power is proportional to cube of velocity. \[ P\propto v^3 \] So: \[ \frac{P'}{P}=\left(\frac{v'}{v}\right)^3 \] \[ \frac{P'}{P}=(2)^3 \] \[ \frac{P'}{P}=8 \] Therefore, power must be increased by a factor of: \[ \boxed{8} \]