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A canal is 300cm wide and 120cm deep. The water in the canal is flowing with a speed of 20km/hr . How much area will it irrigate in 20 minutes if 8cm of standing water is required?

Answer
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Hint: We are asked to find the area irrigated by the flowing water in 20 minutes, given 8cm of standing or still water is required. For this, firstly calculate the volume of water that flows in the canal in one hour then using this calculate the amount of water irrigated in 20 minutes.

Complete step by step solution:
According to the question we are given,
A canal of width=300cm=3m (as 1m=100cm )
and depth(height)=120cm=1.2m
and water is flowing in the canal with speed =20km/hr
Now, as all our units is in metres we will convert the speed also from km/hr to m/hr,
Thus speed=20km/hr=20,000m/hr (as 1km=1000m )
Now, we have to calculate the volume of water that flows in the canal in one hour,
So we have the formula to calculate volume as,
Volume=Area ×speed
Volume = width of canal×height of canal×speed of water in canal
Substituting the values of width, height and speed we get,
Volume=3m×1.2m×20,000m
Volume=72,000m3
Thus we got the volume of water flowing in the canal in 1 hour or 60 minutes as 72,000m3
Now, we have to calculate volume of water flowing in 20 minutes,
 So, first calculate volume of water flowing in 1 minute =72,00060 (as 1hour=60min )
Volume of water flowing in 20 minutes =72,00060×20=24,000m3
Thus we got the volume of water flowing in the canal in 20 minutes as 24,000m3
Now, we have to find the area covered in 20 minutes when 8cm of standing water is required,
So 8cm=0.08m (as 1m=100cm )
So, now area covered 20minutes with 0.08m of standing water =24,000m30.08m
3,00,000m2
=30hectares (as 1hectare=10,000m2 )

Thus we got the area covered in 20 minutes with 8cm of standing water as 30hectares.

Note: The most important thing to keep in mind while solving such problems is the units. Carefully examine the units of the quantities mentioned in the question and then accordingly convert all the quantities in the same unit. While solving the problem all the quantities should be in the same unit.
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