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By increasing the volumeof water coming outper secondthe water pumpto n times , what is the powerof motorhas to beincreases ? |
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Answer» Solution :Power `P = W/t =(1/2mv^(2))/t` Where m = mass of WATER coming out from pump . ` :. P =(mv^(2))/(2t)` ` =(Vrhov^(2))/(2t) "" [ :. m = rhoV] ` ` :. P = (Alrhot^(2))/(2t) "" [ :. V = Al] ` V = volumeof water coming from the pumpin time t A = CROSS sectional area of pipe ( is constant ) l = length of FLOW of water `v = l/t` velocity of water thrown by pump t = density of water ` :. P = 1/2 A rhov^(2) xx1/t` ` =1/2 Arhov^(3) "" [ :. l/t =v] ` ` :. P = 1/2 Arho v^(3)` , here A and `rho `are constant ` :. P prop v^(3) ""....(1)` ` rArr`VOLUME of water coming from cross SECTION of pipe of length l per second `V = V/t =(Al)/t` ` :. V = Av - to [l/t = v] ` If by increasing the value of V to n times then V = Av (here A is constant ) ` :. ` The value of v has to done to n times so from equation (1) , ` :. ` The value of v has to done t n times so from equation (1) , New power of pump , `P prop (nv)^(3)` ` :. P prop n^(3) v^(3)` ` :. P prop n^(3) P ` |
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