# Please solve q 69?

Jun 25, 2018

#### Answer:

The answer is $o p t i o n \left(c\right)$

#### Explanation:

The speed of water in the pipe is $= v$

The diameter of the pipe is $= d$

The volumetric rate flow is $Q = \frac{\pi}{4} {d}^{2} v$

The power is

$\text{ Power } = \Delta P Q$

where,

$\Delta P$ is the pressure drop.

$\Delta P = \left(2 f \left(\frac{l}{d}\right) \cdot \rho {v}^{2}\right)$

Therefore,

$\text{ Power } = \Delta P Q = \left(2 f \left(\frac{l}{d}\right) \cdot \rho {v}^{2}\right) \cdot \left(\frac{\pi}{4} {d}^{2} v\right)$

$= k {v}^{3}$

The answer is $o p t i o n \left(c\right)$

Jun 27, 2018

(c)

#### Explanation:

Alternate method.

Let $A$ be area of cross section of the pipe through which water flows.
Quantity $Q$ of the water flowing through area of cross section in unit time is

$Q = A v$ .....(1)

We know that power $P$ is rate of doing work. Let $m$ be mass of water of quantity $Q$, work done is the kinetic energy of the water

$\text{Work done} = \frac{1}{2} m {v}^{2}$ ........(2)

Since $m \propto Q$, we can write (2) as

$\text{Work done per unit time} \propto Q {v}^{2}$

Inserting value of $Q$ from (2) we get

$\text{Work done per unit time} \propto \left(A v\right) {v}^{2}$
$\implies P \propto {v}^{3}$