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A pump motor is used to deliver water at a certain rate from a given pipe. To obtain `n` times water from the same pipe same time, the factor by which the power of the motor should be increased is: a. n² b. n³c. n4 d. n½

ashu Joshi , 10 Years ago
Grade 12
anser 4 Answers
Arun Kumar

Last Activity: 10 Years ago

power = F.v

v has to be increased by n
oppossing force is proportional to v^2

totally n^3


ravi yadav

Last Activity: 7 Years ago

mass of water flowing volume density per second==(volume X density)/time 
=(area x lenght x density)/time 
=area x density x velocity 
=a z x v 
Therefore,Rate of increase of k.e=mv2/2xtime 
=(A x z x V^3)/2 
Mass m, flowing out per second, can be increased to m1 by increasing to V to V1, then power 
increases from P to P1. 
therefore,p1/p=(A x z x V1^3)/(A x z x V^3) 
=(v1/v)^3 
so, 
m1/m=(A x z x v1)/(a x z x v)=v1/v 
But given m1 = 2m, v1 = 2v 
therefore,p1/p=2^3=8 
so,p1=8p.. 

Siddhant vikram awasthi

Last Activity: 7 Years ago

There is increase in the rate of water flowing through the pipe , sodmdt=Avρwhere A is area−crossection, v is the velocity through which water is flowing out of the pipe , ρ is the density of the materialdm`dt=ndmdtAv`ρ=nAvρ , so v` = nv(a) Force=dpdt=dmvdt=vdmdt=F`F=v`dm`dtvdmdt= nvndmdtvdmdt=n2 , F` = n2 F(b) Power = F.v =P`P =v`dm`dtv`vdmdtv= n2 .n=n3 ,P`=n3P

Kushagra Madhukar

Last Activity: 4 Years ago

Dear student,
Please find the attached to your problem.
 
We know, discharge, Q = Area(A) x Velocity(v)
To increase Q by n times we need to either increase A or v or both.
Since Area of cross section of pipe is same, velocity must be increased by n times.
Now, the force by which water comes out and is exerted on pipe is, F = d(mv)/dt = v*dm/dt = v * ρAv = ρAv2
Hence power delivered is, P = F.v = ρAv2 * v = ρAv3
Hence, P is directly proportional to cube of velocity.
Thus on increasing v, n times, the power becomes, P’ = ρA(nv)3 = n3 * P
Hence the correct option is b) n3
 
Hope it helps.
Thanks and regards,
Kushagra

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