Water flows through a circular pipe, whose internal diameter is 2 cm, at the rate of 0.7 m per second into a cylindrical tank, the radius of whose base is 40 cm. By how much will the level of water in the cylindrical tank use in half an hour?
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Given diameter of the circular pipe = 2 cm
So, the radius of the circular pipe = 2/2 = 1 cm
Height of the circular pipe = 0.7 m = 0.7*100 = 70 cm
Now, volume of the water flows in 1 second = πr2 h
= 3.142*12 *70
= 3.142 * 70
Volume of the water flows in 1/2 hours = 3.142 * 70*30*60
Now, volume of the water flows = Volume of the cylinder
=> 3.142 * 70*30*60 = πr2 h
=> 3.142 * 70*30*60 = 3.142*(40)2 h
=> 70*30*60 = 40*40* h
=> h = (70*30*60)/(40*40)
=> h = (70*3*6)/(4*4)
=> h = 1260/16
=> h = 78.85 cm
So, the level of water rise in the tank in half an hour is 78.75 cm
Hope it helps you
So, the radius of the circular pipe = 2/2 = 1 cm
Height of the circular pipe = 0.7 m = 0.7*100 = 70 cm
Now, volume of the water flows in 1 second = πr2 h
= 3.142*12 *70
= 3.142 * 70
Volume of the water flows in 1/2 hours = 3.142 * 70*30*60
Now, volume of the water flows = Volume of the cylinder
=> 3.142 * 70*30*60 = πr2 h
=> 3.142 * 70*30*60 = 3.142*(40)2 h
=> 70*30*60 = 40*40* h
=> h = (70*30*60)/(40*40)
=> h = (70*3*6)/(4*4)
=> h = 1260/16
=> h = 78.85 cm
So, the level of water rise in the tank in half an hour is 78.75 cm
Hope it helps you
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