Question

A pump on the ground floor of a building can pump up water to fill a tank of volume 30 metre square in 15 minut if the tank is 40 metre above the ground an the efficiency of pump is 30% how much electric power is consumed by the pump.​

Answer 1

Answer:

ANSWER

Here, Volume of water =30 m  

3

 

t=15 min=15×60 s=900 s;Height,h=40 m

Efficiency, η=30%

Density of water =10  

3

kg m  

−3

 

∴ Mass of water pumped =Volume×Density

=(30 m  

3

)(10  

3

 kg m  

−3

)=3×10  

4

kg

P  

output

​  

=  

t

W

​  

=  

t

mgh

​  

=  

900 s

(3×10  

4

kg)(10 ms  

−2

)(40 m)

​  

 

=  

3

4

​  

×10  

4

W

Efficiency, η=  

P  

input

​  

 

P  

output

​  

 

​  

 

P  

input

​  

=  

η

P  

output

​  

 

​  

=  

3×  

100

30

​  

 

4×10  

4

 

​  

=  

9

4

​  

×10  

5

 

=44.4×10  

3

W=44.4 kW.

Explanation:

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Answer 2

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Answer:-

4.0

22

Answer:

4.05 kWh

Given that, A pump can fill a tank of 27 cm³ which is at a height of 40 m in 20 minutes. Also, The efficiency of the pump is 40%.

And, We have to find the electric power consumed by the pump in that process.

First, Let's find the work done which equal to change in potential energy ( because the pump has to work against gravity to pull water of 27m³ upto an height of 40 m )

⇒ Work Done = P E

⇒ Work Done = mgh

Because we don't know the mass of tank when it is completely filled with water, Let's apply the following formula here,

⇒ Density = Mass / Volume

Density of water = 1000 kg/m³

⇒ 1000 = Mass / 27 cm³

converting volume into m³,

⇒ 1000 = Mass / 2.7 × 10⁻⁵

⇒ Mass = 270 × 10⁻⁵ kg

Hence,

⇒ Work done = 270 × 10⁻⁵ × 10 × 40

g = 10 m/s², h = 40 m

Divide it by 1000 to convert it into kiloJoule (kJ)

⇒ Work Done = 27 × 4 × 10³ × 10⁻⁵

⇒ Work Done = 108 × 10⁻²

⇒ Work done = 1.08 kJ

Now, we know the time taken to pull 27000 kg of water upto an height of 40 m to be 20 minutes, Converting it into hour we get 2/3 hour

As we know,

⇒ Power = Work Done / time taken

⇒ Power = 1.08 / 2/3

⇒ Power = 1.62 kWh

Given the efficiency of the water pump is 40%,

⇒ Efficiency = Output / Input

(In terms of power)

⇒ 40/100 = 1.62 / Input

⇒ 0.4 × Input = 1.62

⇒ Input = 4.05 kWh

Hence, The electric power absorbed by the water pump is 4.05 kWh.

__________________________

100% correct answer

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