An electric pump is used to fill an overhead tank of capacity 9 m3 kept at a height of 10 m above the ground. If the pump takes 5 min to fill the tank by consuming 10 kW power the efficiency of the pump should be (Take g=10 text ms-2 )

Question

An electric pump is used to fill an overhead tank of capacity 9 m3 kept at a height of 10 m above the ground. If the pump takes 5 min to fill the tank by consuming 10 kW power the efficiency of the pump should be (Take g=10 text ms-2 ) (A) 60% (B) 40% (C) 20% (D) 30%

Tardigrade Admin · Accepted Answer

Work done in raising water =mgh or W=(volume× density)gh =(9× 1000)× 10× 10 or W=9× 105J ∴ Useful power =(work/t) =(9× 105/5× 60)=3 kW Hence, efficiency =(useful text power/consuming text power) =(3/10)=30%

Q. An electric pump is used to fill an overhead tank of capacity $9 m^{3}$ kept at a height of $10 m$ above the ground. If the pump takes $5 min$ to fill the tank by consuming $10 kW$ power the efficiency of the pump should be (Take $g = 10 m s^{- 2}$ )

60%

40%

20%

30%

50%

Q. An electric pump is used to fill an overhead tank of capacity 9m3 kept at a height of 10m above the ground. If the pump takes 5min to fill the tank by consuming 10kW power the efficiency of the pump should be (Take g=10ms−2 )

60%

40%

20%

30%

50%

Work done in raising water =mgh or W=(volume×density)gh =(9×1000)×10×10 or W=9×105J ∴ Useful power =twork​ =5×609×105​=3kW Hence, efficiency =consumingpowerusefulpower​ =103​=30

Q. An electric pump is used to fill an overhead tank of capacity $9 m^{3}$ kept at a height of $10 m$ above the ground. If the pump takes $5 min$ to fill the tank by consuming $10 kW$ power the efficiency of the pump should be (Take $g = 10 m s^{- 2}$ )