The area of cross section of a pipe is 5.4 centimetres square and the water pump out of it at the rate of 27 kilometre per hour find the volume of water which flows out of the pipe in 1 minute

rate = 27 km/hr

= 2700000 cm / 60 minutes
= 45000 cm/min
so in 1 minute, the water flows 45000 cm

volume = area of base x height
= 5.4 cm^2 x 45000 cm
= 243000 cm^3

As rate= 27km/h

So when we change it into cm it becomes =2700000cm/60mins
= 45000cm/min
So in 1minute 45000cm water is pumped
As the area of base is 5.4cm^2
So volume=area of base × height
5.4×45000
=243000cm^3
Hence,the volume of water pumped in one minute is 243000cm^3.

243

Explain again plz

243 litres

Area of cross section of pipe = 5.4 cm²

Velocity of water = 27km/h
= 27000m/60min
= 2700000cm/60min
= 45000cm/min
Water thrown out of pipe = Volume of water = 5.4cm² × 45000cm/min
= 243000cm³/min
Since, 1cm³=1ml
Therefore, 243000³=243000ml
=243000/1000 L
=243L

To calculate the volume of water that flows out of the pipe in 1 minute, we will need to convert the given rate from kilometer per hour (km/h) to centimeters per minute (cm/min), since the area of cross section is given in centimeters.

To convert km/h to cm/min, we need to use the following conversion factors:

1 kilometer = 100,000 centimeters (since there are 1000 meters in a kilometer and 1 meter = 100 centimeters)
1 hour = 60 minutes

First, let's convert the rate from km/h to cm/min:

27 km/h * (100,000 cm/km) / (60 min/h)

This calculation will give us the flow rate in centimeters per minute.

Now that we have the flow rate in cm/min and the area of cross section of the pipe (5.4 cm^2), we can calculate the volume of water that flows out in 1 minute.

Volume = Flow rate * Area of cross section

Volume = (Flow rate in cm/min) * (Area of cross section in cm^2)

Applying the formula, we have:

Volume = ((27 km/h) * (100,000 cm/km) / (60 min/h)) * (5.4 cm^2)

Calculate this expression to determine the volume of water that flows out of the pipe in 1 minute.

243 litres

No answer is not correct.