Question

Please solve q 69?

Answer 1

The answer is `option (c)`

Explanation:

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

The diameter of the pipe is `=d`

The volumetric rate flow is `Q=pi/4d^2v`

The power is

`" Power "=DeltaPQ`

where,

`Delta P` is the pressure drop.

`DeltaP=(2f(l/d)*rhov^2)`

Therefore,

`" Power "= DeltaPQ =(2f(l/d)*rhov^2)*(pi/4d^2v )`

`=kv^3`

The answer is `option (c)`

Answer 2

(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=Av` .....(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

`"Work done"=1/2mv^2` ........(2)

Since `mpropQ`, we can write (2) as

`"Work done per unit time"propQv^2`

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

`"Work done per unit time"prop(Av)v^2`
`=>Ppropv^3`