Question

1. Find the power of the engine deployed to lift
water at the rate of 5 litre of water in one
second through an average height of 10 m
from a well of depth 10m. Assume 10% of
power of engine is wasted.​

Answer 1

volume of water lift per second = 5litre = 5000 cm³ [ as we know, 1 litre = 1000cm³]

now, mass of water lift per second , m = 5000cm³ × 1g/cm³ [ density of water is 1g/cm³ ]

= 5000g = 5kg

distance travelled by water, h = height + depth = 10m + 10m = 20m

so, power = mgh/t

= 5kg × 9.8m/s² × 20m/1s

= 980 watt

10% of power of engine is wasted.

so, useful power = 90%

now, power of the engine deployed to lift water = 980/90 × 100 = 1088.8 watt

Answer 2

Answer:

The power of the engine deployed to lift  water at the rate of 5 litre of water in one  second through an average height of 10 m  from a well of depth 10m is 1.098 KW.

Explanation:

Let us note down the values given in the question itself.

Volume of water lifted in 1 second is given 5 litre.

Average height  is given 10m.

Depth of the well given 10m.

percentage of the power of the engine wasted is given 10%.

Now,

Total height = Average height + Depth of the well.

                    = 10 + 10

                    =20.

We know that,

Power = Work done ÷ Time.

           = (5 × 9.8 × 20) ÷ 1

           = 980 watt.

As 10 % of power becomes waste.

Therefore, useful power is 90%.

So, power of the engine deployed = (980 × 100)÷ 90.

                                                          = 1088 W

                                                        or 1.098 KW .