### Q.15:- A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m^{3} in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump?

**Answer:-**Volume of the tank, *V* = 30 m^{3}

Time of operation, *t* = 15 min = 15 × 60 = 900 s

Height of the tank, *h* = 40 m

Efficiency of the pump, η = 30 %

Density of water, *ρ* = 10^{3} kg/m^{3}

Mass of water, *m* = *ρ**V* = 30 × 10^{3} kg

Output power can be obtained as:

*P*_{0} = Work done / Time = *mgh* / *t*

= 30 × 10^{3} × 9.8 × 40 / 900 = 13.067 × 10^{3} W

For input power *P*_{i}*,*, efficiency η, is given by the relation:

η = *P*_{0} / *P*_{i} = 30%

*P*_{i} = 13.067 × 100 × 10^{3} / 30

= 0.436 × 10^{5} W = 43.6 kW