# 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. n³ c. n4 d. n½

Arun Kumar IIT Delhi
10 years ago
power = F.v
v has to be increased by n
oppossing force is proportional to v^2

totally n^3

13 Points
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
19 Points
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
4 years ago
Dear student,

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