Please solve q 69?

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2 Answers
Jun 25, 2018

Answer:

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)#

Jun 27, 2018

Answer:

(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#