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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. n³ c. n4 d. n½
power = F.vv has to be increased by noppossing force is proportional to v^2totally n^3
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..
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
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 = ρAv2Hence power delivered is, P = F.v = ρAv2 * v = ρAv3Hence, P is directly proportional to cube of velocity.Thus on increasing v, n times, the power becomes, P’ = ρA(nv)3 = n3 * PHence the correct option is b) n3 Hope it helps.Thanks and regards,Kushagra
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