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If `omega` is unit weight of water, Q the discharge in cumecs, H the total head lift and `eta`, the efficiency of the pump, the H.P. of the motor is


Options:
A .  `H.P = (omega Q H)/( 75 eta)`
B .  `H.P = (omega Q H)/( 4500 eta)`
C .  `H.P = (omega Q eta)/( 75 H)`
D .  `H.P = (omega Q eta)/( 4500 H)`
Answer: Option A


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